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Derivative

Tags

Cegep1

Mathematics

Word count

646 words

Reading time

5 minutes

Slope of a tangent line of a function
Also instantaneous rate of change of the function at the tangent point
The derivative of $f$ at $x$ is

$$f^{\prime }(x)=y^{\prime }=\frac{\mathrm{d}}{\mathrm{d}x}[f(x)]=\frac{\mathrm{d}y}{\mathrm{d}x}=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$$

$\frac{f(x + h) - f(x)}{h}$ is called the **difference quotient**.

$\frac{\mathrm{d}y}{\mathrm{d}x}$ is called the **differentiation operator**, or formally Leibniz's notation, and means "take the derivative with respect to $x$."

+

$f^{\prime }(a)$ stands for derivative at $a$ while $\frac{d}{dx}f(a)$ is derivative at $y=f(a)$, which is always 0.
Under Leibniz's notation we write

$${\frac{d}{dx}f(x)|}_{x=a}$$

Sidedness

Left-side derivative at $a$:

$$f_{-}^{\prime }(a)=\underset{h\to 0^{-}}{lim}\frac{f(a+h)-f(a)}{h}$$

Right-side derivative at $a$:

$$f_{+}^{\prime }(a)=\underset{h\to 0^{+}}{lim}\frac{f(a+h)-f(a)}{h}$$

$f'(x) = L \iff f'_-(x) = f'_+(x) = L$

Order

The $n$ ^th order derivative of $f$ is:

$n$	$f^{(n)}(x)$	$y^{(n)}$	$\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}f(x)$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}$
0	$f(x)$ or $f^{(0)}(x)$	$y$ or $y^{(0)}$	$\frac{{\mathrm{d}}^0}{\mathrm{d}{x}^0}f(x)$	$\frac{\mathrm{d}{y}^0}{\mathrm{d}{x}^0}$
1	$f^{\prime }(x)$ or $f^{(1)}(x)$	$y^{\prime }$ or $y^{(1)}$	$\frac{\mathrm{d}}{\mathrm{d}x}f(x)$	$\frac{\mathrm{d}y}{\mathrm{d}x}$
2	$f^{″}(x)$ or $f^{(2)}(x)$	$y^{″}$ or $y^{(2)}$	$\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}f(x)$	$\frac{\mathrm{d}{y}^2}{\mathrm{d}{x}^2}$
3	$f^{‴}(x)$ or $f^{(3)}(x)$	$y^{‴}$ or $y^{(3)}$	$\frac{{\mathrm{d}}^3}{\mathrm{d}{x}^3}f(x)$	$\frac{\mathrm{d}{y}^3}{\mathrm{d}{x}^3}$
4	$f^{(4)}(x)$	$y^{(4)}$	$\frac{{\mathrm{d}}^4}{\mathrm{d}{x}^4}f(x)$	$\frac{\mathrm{d}{y}^4}{\mathrm{d}{x}^4}$
$⋮$	$⋮$	$⋮$	$⋮$	$⋮$

Properties & theorems

• $\frac{d}{dx}c=0$
• $\frac{d}{dx}(c\cdot f(x))=c\cdot f^{\prime }(x)$
• $\frac{d}{dx}(f(x)\pm g(x))=f^{\prime }(x)\pm g^{\prime }(x)$
• Product Rule: $\frac{d}{dx}(f(x)\cdot g(x))=f(x)g^{\prime }(x)+g(x)f^{\prime }(x)$
• Quotient Rule: $\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)f^{\prime }(x)-f(x)g^{\prime }(x)}{(g(x){)}^2}\text{, provided }g(x)\ne 0$
• Chain Rule: $\frac{\mathrm{d}}{\mathrm{d}x}(f\circ g)(x)=(f^{\prime }\circ g)(x)g^{\prime }(x)$
• General Power Rule: $\frac{d}{dx}f(x{)}^n=nf(x{)}^{n-1}{f}^{\prime }(x)$
• $\frac{d}{dx}\sin x=\cos x$
• $\frac{d}{dx}\cos x=-\sin x$
• $\frac{d}{dx}\tan x={\sec }^{2}x$
• $\frac{d}{dx}\sec x=\tan x\sec x$
• $\frac{d}{dx}\csc x=-\cot x\csc x$
• $\frac{d}{dx}\cot x=-{\csc }^{2}x$
• $\frac{\mathrm{d}}{\mathrm{d}x}\arcsin x=\frac{1}{\sqrt{1-x^2}}$
• $\frac{\mathrm{d}}{\mathrm{d}x}\arccos x=-\frac{1}{\sqrt{1-x^2}}$
• $\frac{\mathrm{d}}{\mathrm{d}x}\arctan x=\frac{1}{1+x^2}$
• $\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arcsec}x=\frac{1}{|x|\sqrt{x^2-1}}$
• $\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arccsc}x=-\frac{1}{|x|\sqrt{x^2-1}}$
• $\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arccot}x=-\frac{1}{1+x^2}$
• $\frac{\mathrm{d}}{\mathrm{d}x}{\log }_{b}x=\frac{1}{x\ln b}$
• $\frac{\mathrm{d}}{\mathrm{d}x}{a}^x=a^x\ln a$

Examples

Find $k^{\prime }(1)$ of $k(x)=(x^2+3{)}^3(x-2x^2{)}^2$.

$$\begin{array}{rl}{k}^{\prime }(1)& ={\frac{\mathrm{d}}{\mathrm{d}x}k(x)|}_{x=1}\\ & ={\frac{\mathrm{d}}{\mathrm{d}x}((x^2+3{)}^3(x-2x^2{)}^2)|}_{x=1}\\ & ={3(x^2+3{)}^2(x^2+3{)}^{\prime }(x-2x^2{)}^2+(x^2+3{)}^{3}2(x-2x^2)(x-2x^2{)}^{\prime }|}_{x=1}\\ & ={3(x^2+3{)}^{2}2x(x-2x^2{)}^2+(x^2+3{)}^{3}2(x-2x^2)(1-4x)|}_{x=1}\\ & =3(4{)}^2\cdot 2(1-2{)}^2+(4{)}^3\cdot 2(1-2)(1-4)\\ & =3\cdot 16\cdot 2\cdot 1+64\cdot 2\cdot -1\cdot -3\\ & =96+384\\ & =480\end{array}$$

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Find the derivative of $h(y)=\tan (\sec ((y^3+1{)}^4))$.

$$\begin{array}{rl}{h}^{\prime }(y)& =\frac{\mathrm{d}}{\mathrm{d}y}h(y)\\ & =\frac{\mathrm{d}}{\mathrm{d}y}(\tan (\sec ((y^3+1{)}^4)))\\ & ={\sec }^2(\sec ((y^3+1{)}^4))(\sec ((y^3+1{)}^4){)}^{\prime }\\ & ={\sec }^2(\sec ((y^3+1{)}^4))\sec ((y^3+1{)}^4)\tan ((y^3+1{)}^4)4(y^3+1{)}^{3}3y^2\end{array}$$

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Find the derivative of $y=\sqrt[4]{x+\sqrt[5]{x+\sqrt[4]{x}}}$.

Let $4rt(x)=\sqrt[4]{x},5rt(x)=\sqrt[5]{x}$,
then $4r{t}^{\prime }(x)=\frac{1}{4}{x}^{1/4-1}=\frac{1}{4}{x}^{-3/4},5r{t}^{\prime }(x)=\frac{1}{5}{x}^{1/5-1}=\frac{1}{5}{x}^{-4/5}$.

$$\begin{array}{rl}{y}^{\prime }& =\frac{\mathrm{d}}{\mathrm{d}x}\sqrt[4]{x+\sqrt[5]{x+\sqrt[4]{x}}}\\ & =\frac{\mathrm{d}}{\mathrm{d}x}4rt(x+5rt(x+4rt(x)))\\ & =4r{t}^{\prime }(x+5rt(x+4rt(x)))\cdot \frac{\mathrm{d}}{\mathrm{d}x}(x+5rt(x+4rt(x)))\\ & =4r{t}^{\prime }(x+5rt(x+4rt(x)))\cdot (1+\frac{\mathrm{d}}{\mathrm{d}x}5rt(x+4rt(x)))\\ & =\frac{1}{4}(x+5rt(x+4rt(x)){)}^{-3/4}\cdot (1+5r{t}^{\prime }(x+4rt(x))\cdot \frac{\mathrm{d}}{\mathrm{d}x}(x+4rt(x)))\\ & =\frac{1}{4}(x+\sqrt[5]{x+\sqrt[4]{x}}{)}^{-3/4}\cdot (1+\frac{1}{5}(x+4rt(x){)}^{-4/5}\cdot (1+4r{t}^{\prime }(x)))\\ & =\frac{1}{4}(x+\sqrt[5]{x+\sqrt[4]{x}}{)}^{-3/4}\cdot (1+\frac{1}{5}(x+\sqrt[4]{x}{)}^{-4/5}\cdot (1+\frac{1}{4}{x}^{-3/4}))\end{array}$$

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