MATH 409 Lecture 14

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Lecture Slides

Part 3: Differential and Integral Calculus

Derivative

A function $f$ is said to be differentiable at a point $a\in \mathbb {R}$ if it is defined on an open interval containing $a$ and the following limit exists

$\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}$

The limit is denoted $f'(a)$ and called the derivative of $f$ at $a$.

An equivalent condition is

$f'(a)=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$

Examples

Constant function. $f(x)=c$ for $x\in \mathbb {R}$.

${\frac {f(x+h)-f(x)}{h}}={\frac {c-c}{h}}=0$ for all $x\in \mathbb {R}$ and $h\neq 0$.

Therefore the limit is 0 and $f$ is differentiable on $\mathbb {R}$ and $f'(x)=0$ for all $x\in \mathbb {R}$.

Identity function. $f(x)=x$ for $x\in \mathbb {R}$.

${\frac {f(x+h)-f(x)}{h}}={\frac {x+h-x}{h}}=1$ for all $x\in \mathbb {R}$, $h\neq 0$.

Therefore $f$ is differentiable on $\mathbb {R}$ and $f'(x)=1$ for all $x\in \mathbb {R}$.

Quadratic function. $f(x)=x^{2}$ for $x\in \mathbb {R}$.

${\frac {f(x+h)-f(x)}{h}}={\frac {(x+h)^{2}-x^{2}}{h}}={\frac {2xh+h^{2}}{h}}=2x+h$.

Therefore $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}(2x+h)=2x$ (limit of continuous function)

Therefore $f$ is differentiable on $\mathbb {R}$ and $f'(x)=2x$ for $x\in \mathbb {R}$.

Harmonic function. $f(x)={\frac {1}{x}}$, $x\in (-\infty ,0)\cup (0,\infty )$.

${\frac {f(x+h)-f(x)}{h}}={\frac {1}{h}}\cdot \left({\frac {1}{x+h}}-{\frac {1}{x}}\right)={\frac {1}{h}}\cdot {\frac {x-(x+h)}{(x+h)\,x}}=-{\frac {1}{(x+h)\,x}}$.

Therefore $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}-{\frac {1}{(x+h)\,x}}=-{\frac {1}{x^{2}}}$.

That is, $f$ is differentiable on $\mathbb {R} \setminus \left\{0\right\}$ and $f'(x)=-{\frac {1}{x^{2}}}$ for all $x\neq 0$.

Square root. $f(x)={\sqrt {x}}$, where $x\in [0,\infty )$.

${\frac {f(x+h)-f(x)}{h}}={\frac {{\sqrt {x+h}}-{\sqrt {x}}}{h}}\cdot {\frac {{\sqrt {x+h}}+{\sqrt {x}}}{{\sqrt {x+h}}+{\sqrt {x}}}}={\frac {(x+h)-(x)}{h({\sqrt {x+h}}+{\sqrt {x}})}}={\frac {1}{{\sqrt {x+h}}+{\sqrt {x}}}}$.

Therefore $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}{\frac {1}{{\sqrt {x+h}}+{\sqrt {x}}}}={\frac {1}{2{\sqrt {x}}}}$.

In the case $x=0$,

$\lim _{h\to 0^{+}}{\frac {f(h)-f(0)}{h}}=\lim _{h\to 0}{\frac {1}{\sqrt {h}}}=+\infty$.

Hence $f$ is differentiable on $(0,\infty )$, and $f'(x)={\frac {1}{2{\sqrt {x}}}}$ for all $x>0$.

Sine function. $f(x)=\sin {x}$ for all $x\in \mathbb {R}$.

Using the formula $\sin {\alpha }-\sin {\beta }=2\,\sin {\frac {\alpha -\beta }{2}}\,\cos {\frac {\alpha +\beta }{2}}$, we obtain

${\frac {f(x+h)-f(x)}{h}}={\frac {\sin {(x+h)}-\sin {x}}{h}}={\frac {2}{h}}\,\sin {\frac {h}{2}}\,\cos {\frac {2x+h}{2}}$

Therefore

${\begin{aligned}\lim _{h\to 0}{\frac {2}{h}}\,\sin {\frac {h}{2}}\,\cos {\frac {2x+h}{2}}&=\lim _{h\to 0}{\frac {\sin {\frac {h}{2}}}{\frac {h}{2}}}\cdot \lim _{h\to 0}\cos {\left(x+{\frac {h}{2}}\right)}\\&=1\cdot \cos {x}\end{aligned}}$

That is, $f$ is differentiable on $\mathbb {R}$ and $f'(x)=\cos {x}$ for all $x\in \mathbb {R}$.

Differentiability Theorems

Differentiability and Continuity

Theorem. If a function $f$ is differentiable at a point $a$, then it is continuous at $a$.

Proof.

${\begin{aligned}\lim _{x\to a}f(a)&=\lim _{x\to a}\left(f(x)+{\frac {f(x)-f(a)}{x-a}}\,(x-a)\right)\\&=\lim _{x\to a}f(a)+\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}\cdot \lim _{x\to a}(x-a)\\&=f(a)+f'(a)\cdot 0=f(a)\end{aligned}}$

quod erat demonstrandum

A simple statement that will be used many times:

Note: Similarly, if $f$ has a right-hand derivative at $a$, then $\lim _{x\to a^{+}}f(x){=}f(a)$.
If $f$ has a left-hand derivative at $a$, then $\lim _{x\to a^{-}}f(x){=}f(a)$.

Sum Rule and Homogeneous Rule

Theorem. [Sum rule]. If functions $f$ and $g$ are differentiable at a point $a\in \mathbb {R}$, then the sum $f+g$ is also differentiable at $a$. Moreover, $(f+g)'(a)=f'(a)+g'(a)$.

Proof.

${\begin{aligned}\lim _{x\to a}{\frac {(f+g)(x)-(f+g)(a)}{x-a}}&=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}+\lim _{x\to a}{\frac {g(x)-g(a)}{x-a}}\\&=f'(a)+g'(a)\end{aligned}}$

quod erat demonstrandum

It's worth noting that if the domains of $f$ and $g$ are not equal, then the domain of the sum of the functions is defined where the functions overlap. The sum over this open interval is well defined.

Theorem. [Homogeneous rule]. If a function $f$ is differentiable at a point $a\in \mathbb {R}$, then for any $r\in \mathbb {R}$ the scalar multiple $r\,f$ is also differentiable at $a$. Moreover, $(r\,f)'(a)=r\,f'(a)$.

Proof.

${\begin{aligned}\lim _{x\to a}{\frac {(r\,f)(x)-(r\,f)(a)}{x-a}}&=\lim _{x\to a}r\,{\frac {f(x)-f(a)}{x-a}}\\&=r\,f'(a)\end{aligned}}$

quod erat demonstrandum

Product Rule

Theorem. [Product rule]. If the functions $f$ and $g$ are differentiable at a point $a\in \mathbb {R}$, then the product $f\cdot g$ is also differentiable at $a$. Moreover, $(f\cdot g)'(a)=f'(a)\,g(a)+f(a)\,g'(a)$.

Proof. Since $f$ and $g$ are differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} , there is an open interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I = (c,d)} containing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} such that both Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are defined on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} . (see note above regarding sums) For every Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in I \setminus \left\{ a \right\}} we have:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} f(x) \, g(x) - f(a) \, g(a) &= f(x) \, g(x) - f(a) \, g(a) + f(a) \, g(x) - f(a) \, g(a) \\ &= (f(x) - f(a)) \, g(x) + f(a) \, (g(x) - g(a)) \end{align}}

Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{(f \cdot g)(x) - (f \cdot g)(a)}{x - a} = \frac{f(x)-f(a)}{x-a} \, g(x) + f(a) \, \frac{g(x)-g(a)}{x-a}} so that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \lim_{x \to a} \frac{(f \cdot g)(x) - (f \cdot g)(a)}{x-a} &= \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \cdot \lim_{x \to a} g(x) + \lim_{x \to a} f(a) \cdot \lim_{x \to a} \frac{g(x)-g(a)}{x-a} \\ &= f'(x) \, g(a) + f(a) \, g'(a) \end{align}}

We used the fact that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are continuous at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} to evaluate the limits in the last step.

quod erat demonstrandum

Reciprocal Rule

Theorem. [Reciprocal rule]. If a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is differentiable at a point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in \mathbb{R}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(a) \ne 0} , then the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{f}} is also differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} . Moreover, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( \frac{1}{f} \right)'(a) = -\frac{f'(a)}{f^2(a)}} .

Proof. The function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is defined on an open interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (c,d)} containing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} . We know that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is continuous at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon = \left| f(a) \right| > 0} , there exists Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta > 0} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| f(x) - f(a) \right| < \epsilon} for any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in I = (c,d) \cap (a - \delta, a + \delta)} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) \ne 0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in I} . In particular, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{f}} is defined on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} , an open interval containing $a$.

Now

${\begin{aligned}\lim _{x\to a}{\frac {\left({\frac {1}{f}}\right)(x)-\left({\frac {1}{f}}\right)(a)}{x-a}}&=\lim _{x\to a}\left({\frac {1}{f(x)}}-{\frac {1}{f(a)}}\right)\,{\frac {1}{x-a}}\\&=\lim _{x\to a}{\frac {f(a)-f(x)}{f(x)\,f(a)}}\,{\frac {1}{x-a}}\\&=\lim _{x\to a}\left(-{\frac {f(x)-f(a)}{x-a}}\,{\frac {1}{f(x)\,f(a)}}\right)\\&=-\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}\,\lim _{x\to a}{\frac {1}{f(x)\,f(a)}}\\&={\frac {f'(a)}{f^{2}(a)}}\end{aligned}}$

quod erat demonstrandum

Difference Rule and Quotient Rule

Theorem. [Difference rule]. If $f$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable at a point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in \mathbb{R}} , then the difference Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f-g)} is also differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} . Moreover, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f-g)'(a) = f'(a) - g'(a)} .

Proof. By the Homogeneous Rule, the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -g = (-1) \, g} is differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-g)'(a) = -g'(a)} By the Sum Rule, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f-g = f+ (-g)} is also differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f-g)'(a) = f'(a) + (-g)'(a) = f'(a) - g'(a)} .

quod erat demonstrandum

Theorem. [Quotient rule]. If functions Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and $g$ are differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in \mathbb{R}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(a) \ne 0} , then the quotient Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{f}{g}} is also differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} . Moreover, $\left({\frac {f}{g}}\right)'(a)={\frac {f'(a)\,g(a)-f(a)\,g'(a)}{g^{2}(a)}}$.

Proof. By the Reciprocal Rule, the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{g}} is differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{1}{g}\right)'(a) = \frac{-g'(a)}{g^2(a)}} . By the product rule, the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{f}{g} = f \cdot \frac{1}{g}} is also differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \left( \frac{f}{g} \right)'(a) &= \frac{f'(a)}{g(a)} + f(a) \, \left( \frac{1}{g} \right)'(a) \\ &= \frac{f'(a)\,g(a) - f(a) \, g'(a)}{g^2(a)} \end{align}}

quod erat demonstrandum

Chain Rule

Theorem. [Chain rule]. If a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is differentiable at a point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in \mathbb{R}} and a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(a)} , then the composition Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g \circ f} is differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} . Moreover, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (g \circ f)'(a) = g'(f(a)) \cdot f'(a)} .

Proof. The function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is defined on an open interal Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I = (a - \delta, a + \delta)} while Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is defined on an open interal Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J = (f(a) - \epsilon, f(a) + \epsilon)} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is continuous at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} (because it is differentiable), there exists Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_0 \in (0, \delta)} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(I_0) \subset J} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_0 = (a-\delta_0, a + \delta_0)} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g \circ f} is defined on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_0} . For any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in I_0} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) \ne f(a)} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{(g \circ f)(x) - (g \circ f)(a)}{x-a} = \frac{g(f(x)) - g(f(a))}{f(x) - f(a)} \cdot \frac{f(x) - f(a)}{x-a}}

This implies the Chain Rule unless there is a sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ x_n \right\}} converging to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_n \ne a} while Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x_n) = f(a)} . In this case, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (g \circ f)'(a) = f'(a) = 0} .

quod erat demonstrandum

Examples

Cosine function. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \cos{x}} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \mathbb{R}} .

The function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} can be represented as a composition Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f = h \circ g} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) = x + \frac{\pi}{2}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x) = \sin{x}} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \mathbb{R}} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x) = 1} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x) = \cos{x}} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \mathbb{R}} , the Chain rule implies that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = h'(g(x)) \, g(x) = \cos{\left( x + \frac{\pi}{2} \right)} = -\sin{x}} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \mathbb{R}} .

Tangent function. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \tan{x}} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)} .

Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \frac{\sin{x}}{\cos{x}}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{x} \ne 0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)} , the Quotient Rule implies that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)} and

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = \frac{(\sin{x})' \, \cos{x} - \sin{x} \, (\cos{x})'}{\cos^2{x}} = \frac{\cos^2{x} + \sin^2{x}}{\cos^2{x}} = \frac{1}{\cos^2{x}}}

For all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)} .