 maths > differential-calculus

Derivatives of Algebraic Expressions

what you'll learn...

Overview

→  standard functions in algebraic expressions

→  trigonometric functions sin, cos, tan etc.

→  inverse trigonometric functions such as $\mathrm{arcsin}$$\arcsin$, $\mathrm{arccos}$$\arccos$, etc.

→  exponents and logarithmic functions such as ${e}^{x}$${e}^{x}$, ${a}^{x}$${a}^{x}$, and $\mathrm{ln}x$$\ln x$

constant

Finding the derivative of $y=a$$y = a$ in first principles:

$\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$

$=\underset{\delta \to 0}{lim}\left[f\left(x+\delta \right)$$-f\left(x\right)\right]/\delta$

$=\underset{\delta \to 0}{lim}\left[a$$-a\right]/\delta$

$=\underset{\delta \to 0}{lim}\left[0\right]/\delta$$= {\lim}_{\delta \to 0} \left[0\right] / \delta$

The above proves
$\frac{d}{dx}\left(a\right)=0$$\frac{d}{\mathrm{dx}} \left(a\right) = 0$
Note that $\delta$$\delta$ is very small value and the numerator is $0$$0$. So the limit evaluates to $0$$0$.

exponent

Finding the derivative of $y={x}^{n}$$y = {x}^{n}$ in first principles:

$\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$

$=\underset{\delta \to 0}{lim}\left[f\left(x+\delta \right)$$-f\left(x\right)\right]/\delta$

$=\underset{\delta \to 0}{lim}\left[{\left(x+\delta \right)}^{n}$$-{x}^{n}\right]/\delta$

using binomial expression for ${\left(x+\delta \right)}^{n}$${\left(x + \delta\right)}^{n}$ $=\underset{\delta \to 0}{lim}\left[\left({x}^{n}+n\delta {x}^{n-1}$${+}^{n}{C}_{2}{\delta }^{2}{x}^{n-2}$${+}^{n} {C}_{2} {\delta}^{2} {x}^{n - 2}$$+\cdots +{\delta }^{n}\right)$$-{x}^{n}\right]/\delta$

canceling ${x}^{n}$$\textcolor{c \mathmr{and} a l}{{x}^{n}}$ and $-{x}^{n}$$\textcolor{\mathrm{de} e p s k y b l u e}{- {x}^{n}}$
$=\underset{\delta \to 0}{lim}\left[\left(n\delta {x}^{n-1}$${+}^{n}{C}_{2}{\delta }^{2}{x}^{n-2}$${+}^{n} {C}_{2} {\delta}^{2} {x}^{n - 2}$$+\cdots +{\delta }^{n}\right)$$\right]/\delta$

dividing all terms by $\delta$$\delta$
$=\underset{\delta \to 0}{lim}\left[\left(n{x}^{n-1}$${+}^{n}{C}_{2}{\delta }^{1}{x}^{n-2}$${+}^{n} {C}_{2} {\delta}^{1} {x}^{n - 2}$$+\cdots +{\delta }^{n-1}\right)$$\right]$

applying limit $\delta \to 0$$\delta \to 0$
The $\delta$$\delta$ is substituted as $0$$0$ and only one term is non zero. The above proves
$\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$".

summary

Derivatives of Algebraic Expressions :
Derivative of a constant
$\frac{d}{dx}a=0$$\frac{d}{\mathrm{dx}} a = 0$
rate of change of constant is 0

Derivative of power:
$\frac{d}{dx}{x}^{n}=n{x}^{n-1}$$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$
small change results in binomial expansion and only one factor remains

example

Find the derivative of ${x}^{3}+4{x}^{2}-3x-2$${x}^{3} + 4 {x}^{2} - 3 x - 2$
The answer is "$3{x}^{2}+8x-3$$3 {x}^{2} + 8 x - 3$"

sine

Finding the derivative of $y=\mathrm{sin}x$$y = \sin x$ in first principles:

$\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$

$=\underset{\delta \to 0}{lim}\left[f\left(x+\delta \right)$$-f\left(x\right)\right]/\delta$

$=\underset{\delta \to 0}{lim}\left[\mathrm{sin}\left(x+\delta \right)$$-\mathrm{sin}x\right]/\delta$

$=\underset{\delta \to 0}{lim}\left[\mathrm{sin}x\mathrm{cos}\delta +\mathrm{cos}x\mathrm{sin}\delta$$-\mathrm{sin}x\right]/\delta$

$=\underset{\delta \to 0}{lim}\frac{\mathrm{cos}x\mathrm{sin}\delta }{\delta }$$= {\lim}_{\delta \to 0} \frac{\textcolor{c \mathmr{and} a l}{\cos x \sin \delta}}{\delta}$$-\underset{\delta \to 0}{lim}\frac{\mathrm{sin}x-\mathrm{sin}x\mathrm{cos}\delta }{\delta }$$- {\lim}_{\delta \to 0} \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\sin x} - \textcolor{c \mathmr{and} a l}{\sin x \cos \delta}}{\delta}$

$=\mathrm{cos}x\underset{\delta \to 0}{lim}\frac{\mathrm{sin}\delta }{\delta }$$= \cos x {\lim}_{\delta \to 0} \frac{\sin \delta}{\delta}$$-\mathrm{sin}x\underset{\delta \to 0}{lim}\frac{1-\mathrm{cos}\delta }{\delta }$$- \sin x {\lim}_{\delta \to 0} \frac{1 - \cos \delta}{\delta}$

applying the standard limits
$=\mathrm{cos}x×1-\mathrm{sin}x×0$$= \cos x \times 1 - \sin x \times 0$

The above proves
$\frac{d}{dx}\mathrm{sin}x=\mathrm{cos}x$$\frac{d}{\mathrm{dx}} \sin x = \cos x$

cosine

Finding the derivative of $y=\mathrm{cos}x$$y = \cos x$:

$\frac{d}{dx}\mathrm{cos}x$$\frac{d}{\mathrm{dx}} \cos x$

$=\frac{d}{dx}\mathrm{sin}\left(\frac{\pi }{2}-x\right)$$= \frac{d}{\mathrm{dx}} \sin \left(\frac{\pi}{2} - x\right)$

applying chain rule of differentiation with $u=\frac{\pi }{2}-x$$u = \frac{\pi}{2} - x$
$=\frac{d}{du}\mathrm{sin}\left(u\right)\frac{d}{dx}\left(\frac{\pi }{2}-x\right)$$= \frac{d}{\mathrm{du}} \sin \left(u\right) \frac{d}{\mathrm{dx}} \left(\frac{\pi}{2} - x\right)$

$=\mathrm{cos}\left(u\right)×\left(-1\right)$$= \cos \left(u\right) \times \left(- 1\right)$

$=-\mathrm{cos}\left(\frac{\pi }{2}-x\right)$$= - \cos \left(\frac{\pi}{2} - x\right)$

$=-\mathrm{sin}\left(x\right)$$= - \sin \left(x\right)$

The above proves
$\frac{d}{dx}\mathrm{cos}x=-\mathrm{sin}x$$\frac{d}{\mathrm{dx}} \cos x = - \sin x$

tangent

Finding the derivative of $y=\mathrm{tan}x$$y = \tan x$

$\left(\mathrm{tan}x\right)\prime$

$=\left(\frac{\mathrm{sin}x}{\mathrm{cos}x}\right)\prime$

applying quotient rule of differentiation
$=\frac{\mathrm{cos}x\left(\mathrm{sin}x\right)\prime -\mathrm{sin}x\left(\mathrm{cos}x\right)\prime }{{\mathrm{cos}}^{2}x}$

$=\frac{\mathrm{cos}x\left(\mathrm{cos}x\right)-\mathrm{sin}x\left(-\mathrm{sin}x\right)}{{\mathrm{cos}}^{2}x}$$= \frac{\cos x \left(\cos x\right) - \sin x \left(- \sin x\right)}{{\cos}^{2} x}$

$=\frac{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}$$= \frac{{\cos}^{2} x + {\sin}^{2} x}{{\cos}^{2} x}$

$=\frac{1}{{\mathrm{cos}}^{2}x}$$= \frac{1}{{\cos}^{2} x}$

$={\mathrm{sec}}^{2}\left(x\right)$$= {\sec}^{2} \left(x\right)$

The above proves
$\frac{d}{dx}\mathrm{tan}x={\mathrm{sec}}^{2}x$$\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$

cotangent

Finding the derivative of $y=\mathrm{cot}x$$y = \cot x$

$\left(\mathrm{cot}x\right)\prime$

$=\left(\frac{\mathrm{cos}x}{\mathrm{sin}x}\right)\prime$

applying quotient rule of differentiation
$=\frac{\mathrm{sin}x\left(\mathrm{cos}x\right)\prime -\mathrm{cos}x\left(\mathrm{sin}x\right)\prime }{{\mathrm{sin}}^{2}x}$

$=\frac{\mathrm{sin}x\left(-\mathrm{sin}x\right)-\mathrm{cos}x\left(\mathrm{cos}x\right)}{{\mathrm{sin}}^{2}x}$$= \frac{\sin x \left(- \sin x\right) - \cos x \left(\cos x\right)}{{\sin}^{2} x}$

$=\frac{-{\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x}$$= \frac{- {\sin}^{2} x - {\cos}^{2} x}{{\sin}^{2} x}$

$=\frac{-1}{{\mathrm{sin}}^{2}x}$$= \frac{- 1}{{\sin}^{2} x}$

$=-{\mathrm{csc}}^{2}\left(x\right)$$= - {\csc}^{2} \left(x\right)$

The above proves
$\frac{d}{dx}\mathrm{cot}x=-{\mathrm{csc}}^{2}x$$\frac{d}{\mathrm{dx}} \cot x = - {\csc}^{2} x$

secant

Finding the derivative of $y=\mathrm{sec}x$$y = \sec x$

$\left(\mathrm{sec}x\right)\prime$

$=\left(\frac{1}{\mathrm{cos}x}\right)\prime$

applying quotient rule of differentiation
$=\frac{\mathrm{cos}x\left(1\right)\prime -\left(\mathrm{cos}x\right)\prime }{{\mathrm{cos}}^{2}x}$

$=\frac{\mathrm{cos}x\left(0\right)-\left(-\mathrm{sin}x\right)}{{\mathrm{cos}}^{2}x}$$= \frac{\cos x \left(0\right) - \left(- \sin x\right)}{{\cos}^{2} x}$

$=\frac{\mathrm{sin}x}{{\mathrm{cos}}^{2}x}$$= \frac{\sin x}{{\cos}^{2} x}$

$=\mathrm{sec}x\mathrm{tan}x$$= \sec x \tan x$

The above proves
$\frac{d}{dx}\mathrm{sec}x=\mathrm{sec}x\mathrm{tan}x$$\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x$

cosecant

Finding the derivative of $y=\mathrm{csc}x$$y = \csc x$

$\left(\mathrm{csc}x\right)\prime$

$=\left(\frac{1}{\mathrm{sin}x}\right)\prime$

applying quotient rule of differentiation
$=\frac{\mathrm{sin}x\left(1\right)\prime -\left(\mathrm{sin}x\right)\prime }{{\mathrm{sin}}^{2}x}$

$=\frac{\mathrm{sin}x\left(0\right)-\left(\mathrm{cos}x\right)}{{\mathrm{sin}}^{2}x}$$= \frac{\sin x \left(0\right) - \left(\cos x\right)}{{\sin}^{2} x}$

$=\frac{-\mathrm{cos}x}{{\mathrm{sin}}^{2}x}$$= \frac{- \cos x}{{\sin}^{2} x}$

$=-\mathrm{csc}x\mathrm{cot}x$$= - \csc x \cot x$

The above proves
$\frac{d}{dx}\mathrm{csc}x=-\mathrm{csc}x\mathrm{cot}x$$\frac{d}{\mathrm{dx}} \csc x = - \csc x \cot x$

summary

Derivatives of Trigonometric Functions:
$\frac{d}{dx}\mathrm{sin}x=\mathrm{cos}x$$\frac{d}{\mathrm{dx}} \sin x = \cos x$

$\frac{d}{dx}\mathrm{cos}x=-\mathrm{sin}x$$\frac{d}{\mathrm{dx}} \cos x = - \sin x$
$\mathrm{cos}x=\mathrm{sin}\left(\frac{\pi }{2}-x\right)$$\cos x = \sin \left(\frac{\pi}{2} - x\right)$ and the negative sign of $x$$x$ is carried in the result

$\frac{d}{dx}\mathrm{tan}x={\mathrm{sec}}^{2}x$$\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$
$\mathrm{tan}=\frac{\mathrm{sin}}{\mathrm{cos}}$$\tan = \frac{\sin}{\cos}$ and results in $\frac{1}{{\mathrm{cos}}^{2}}$$\frac{1}{{\cos}^{2}}$

$\frac{d}{dx}\mathrm{cot}x=-{\mathrm{csc}}^{2}x$$\frac{d}{\mathrm{dx}} \cot x = - {\csc}^{2} x$
$\mathrm{cot}=\mathrm{tan}\left(\frac{\pi }{2}-x\right)$$\cot = \tan \left(\frac{\pi}{2} - x\right)$ and the negative sign of x is carried

$\frac{d}{dx}\mathrm{sec}x=\mathrm{sec}x\mathrm{tan}x$$\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x$
$\mathrm{sec}=\frac{1}{\mathrm{cos}}$$\sec = \frac{1}{\cos}$ and results in $\frac{\mathrm{sin}}{{\mathrm{cos}}^{2}}$$\frac{\sin}{{\cos}^{2}}$

$\frac{d}{dx}\mathrm{csc}x=-\mathrm{csc}x\mathrm{cot}x$$\frac{d}{\mathrm{dx}} \csc x = - \csc x \cot x$
$\mathrm{csc}=\mathrm{sec}\left(\frac{\pi }{2}-x\right)$$\csc = \sec \left(\frac{\pi}{2} - x\right)$ and the negative sign of x is carried

example

Find the derivative of $\mathrm{tan}x-x$$\tan x - x$
The answer is "${\mathrm{tan}}^{2}x$${\tan}^{2} x$".

$\mathrm{tan}x-x$$\tan x - x$
${\mathrm{sec}}^{2}x-1$${\sec}^{2} x - 1$
${\mathrm{tan}}^{2}x$${\tan}^{2} x$

inverse sine

Finding the derivative of $y=\mathrm{arcsin}x$$y = \arcsin x$:
$y=\mathrm{arcsin}x$$y = \arcsin x$
$\mathrm{sin}y=x$$\sin y = x$
differentiate this
$\mathrm{cos}y\frac{dy}{dx}=1$$\cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{dy}{dx}=\frac{1}{\mathrm{cos}y}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos y}$
$=\frac{1}{\sqrt{1-{\mathrm{sin}}^{2}y}}$$= \frac{1}{\sqrt{1 - {\sin}^{2} y}}$
$=\frac{1}{\sqrt{1-{x}^{2}}}$$= \frac{1}{\sqrt{1 - {x}^{2}}}$

The above proves
$\frac{d}{dx}\mathrm{arcsin}x=\frac{1}{\sqrt{1-{x}^{2}}}$$\frac{d}{\mathrm{dx}} \arcsin x = \frac{1}{\sqrt{1 - {x}^{2}}}$

inverse cosine

Finding the derivative of $y=\mathrm{arccos}x$$y = \arccos x$:
$y=\mathrm{arccos}x$$y = \arccos x$
$\mathrm{cos}y=x$$\cos y = x$
differentiate this
$-\mathrm{sin}y\frac{dy}{dx}=1$$- \sin y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{dy}{dx}=\frac{1}{-\mathrm{sin}y}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{- \sin y}$
$\frac{dy}{dx}=\frac{-1}{\sqrt{1-{\mathrm{cos}}^{2}y}}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{\sqrt{1 - {\cos}^{2} y}}$
$\frac{dy}{dx}=\frac{-1}{\sqrt{1-{x}^{2}}}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{\sqrt{1 - {x}^{2}}}$

The above proves
$\frac{d}{dx}\mathrm{arccos}x=\frac{-1}{\sqrt{1-{x}^{2}}}$$\frac{d}{\mathrm{dx}} \arccos x = \frac{- 1}{\sqrt{1 - {x}^{2}}}$

inverse tangent

Finding the derivative of $y=\mathrm{arctan}x$$y = \arctan x$:
$y=\mathrm{arctan}x$$y = \arctan x$
$\mathrm{tan}y=x$$\tan y = x$
differentiate this
${\mathrm{sec}}^{2}y\frac{dy}{dx}=1$${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{dy}{dx}=\frac{1}{{\mathrm{sec}}^{2}y}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} y}$
$=\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}y}}$$= \frac{1}{\sqrt{1 + {\tan}^{2} y}}$
$=\frac{1}{\sqrt{1+{x}^{2}}}$$= \frac{1}{\sqrt{1 + {x}^{2}}}$

The above proves
$\frac{d}{dx}\mathrm{arctan}x=\frac{1}{1+{x}^{2}}$$\frac{d}{\mathrm{dx}} \arctan x = \frac{1}{1 + {x}^{2}}$

inverse secant

Finding the derivative of $y=arc\mathrm{sec}x$$y = a r c \sec x$:
$y=arc\mathrm{sec}x$$y = a r c \sec x$
$y$$y$ is in first or second quadrant.
$\mathrm{sec}y=x$$\sec y = x$
differentiate this
$\mathrm{sec}y\mathrm{tan}y\frac{dy}{dx}=1$$\sec y \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\frac{dy}{dx}=\frac{1}{\mathrm{sec}y\mathrm{tan}y}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec y \tan y}$
$\mathrm{sec}y\mathrm{tan}y$$\sec y \tan y$ is always positive for values in first and second quadrant.
$\frac{dy}{dx}=\frac{1}{\mathrm{sec}y\sqrt{{\mathrm{sec}}^{2}y-1}}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec y \sqrt{{\sec}^{2} y - 1}}$
$=\frac{1}{|x|\sqrt{{x}^{2}-1}}$$= \frac{1}{| x | \sqrt{{x}^{2} - 1}}$

The above proves
$\frac{d}{dx}arc\mathrm{sec}x=\frac{1}{|x|\sqrt{{x}^{2}-1}}$$\frac{d}{\mathrm{dx}} a r c \sec x = \frac{1}{| x | \sqrt{{x}^{2} - 1}}$

inverse cosecant

Finding the derivative of $y=arc\mathrm{csc}x$$y = a r c \csc x$:
$y=arc\mathrm{csc}x$$y = a r c \csc x$
$y$$y$ is in range $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$$- \frac{\pi}{2} \le y \le \frac{\pi}{2}$.
$\mathrm{csc}y=x$$\csc y = x$
differentiate this
$-\mathrm{csc}y\mathrm{cot}y\frac{dy}{dx}=1$$- \csc y \cot y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{dy}{dx}=\frac{-1}{\mathrm{csc}y\mathrm{cot}y}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{\csc y \cot y}$
$\mathrm{csc}y\mathrm{cot}y$$\csc y \cot y$ is always positive for values in range $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$$- \frac{\pi}{2} \le y \le \frac{\pi}{2}$
$=\frac{-1}{\mathrm{csc}y\sqrt{{\mathrm{csc}}^{2}y-1}}$$= \frac{- 1}{\csc y \sqrt{{\csc}^{2} y - 1}}$
$=\frac{-1}{|x|\sqrt{{x}^{2}-1}}$$= \frac{- 1}{| x | \sqrt{{x}^{2} - 1}}$

The above proves
$\frac{d}{dx}arc\mathrm{csc}x=\frac{-1}{|x|\sqrt{{x}^{2}-1}}$$\frac{d}{\mathrm{dx}} a r c \csc x = \frac{- 1}{| x | \sqrt{{x}^{2} - 1}}$

inverse cotangent

Finding the derivative of $y=arc\mathrm{cot}x$$y = a r c \cot x$:
$y=arc\mathrm{cot}x$$y = a r c \cot x$
$\mathrm{cot}y=x$$\cot y = x$
differentiate this
$-{\mathrm{csc}}^{2}y\frac{dy}{dx}=1$$- {\csc}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{dy}{dx}=\frac{-1}{{\mathrm{csc}}^{2}y}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{{\csc}^{2} y}$
$=\frac{-1}{\sqrt{1+{\mathrm{tan}}^{2}y}}$$= \frac{- 1}{\sqrt{1 + {\tan}^{2} y}}$
$=\frac{-1}{\sqrt{1+{x}^{2}}}$$= \frac{- 1}{\sqrt{1 + {x}^{2}}}$

The above proves
$\frac{d}{dx}arc\mathrm{cot}x=\frac{-1}{1+{x}^{2}}$$\frac{d}{\mathrm{dx}} a r c \cot x = \frac{- 1}{1 + {x}^{2}}$

summary

Derivatives of inverse trigonometric functions : $\text{trig fn}\left(y\right)=x$$\textrm{t r i g f n} \left(y\right) = x$, so derivative in terms of x

$\frac{d}{dx}\mathrm{arcsin}x=\frac{1}{\sqrt{1-{x}^{2}}}$$\frac{d}{\mathrm{dx}} \arcsin x = \frac{1}{\sqrt{1 - {x}^{2}}}$

$\frac{d}{dx}\mathrm{arccos}x=\frac{-1}{\sqrt{1-{x}^{2}}}$$\frac{d}{\mathrm{dx}} \arccos x = \frac{- 1}{\sqrt{1 - {x}^{2}}}$

$\frac{d}{dx}\mathrm{arctan}x=\frac{1}{1+{x}^{2}}$$\frac{d}{\mathrm{dx}} \arctan x = \frac{1}{1 + {x}^{2}}$

$\frac{d}{dx}arc\mathrm{sec}x=\frac{1}{|x|\sqrt{{x}^{2}-1}}$$\frac{d}{\mathrm{dx}} a r c \sec x = \frac{1}{| x | \sqrt{{x}^{2} - 1}}$

$\frac{d}{dx}arc\mathrm{csc}x=\frac{-1}{|x|\sqrt{{x}^{2}-1}}$$\frac{d}{\mathrm{dx}} a r c \csc x = \frac{- 1}{| x | \sqrt{{x}^{2} - 1}}$

$\frac{d}{dx}arc\mathrm{cot}x=\frac{-1}{1+{x}^{2}}$$\frac{d}{\mathrm{dx}} a r c \cot x = \frac{- 1}{1 + {x}^{2}}$

example

Find the derivative of $\left(1+{x}^{2}\right)\mathrm{arctan}x$$\left(1 + {x}^{2}\right) \arctan x$
The answer is "$1+2x\mathrm{arctan}x$$1 + 2 x \arctan x$"

$\left[\left(1+{x}^{2}\right)\mathrm{arctan}x\right]\prime$
$\left(1+{x}^{2}\right)\left(\mathrm{arctan}x\right)\prime +\left(1+{x}^{2}\right)\prime \mathrm{arctan}x$
$\left(1+{x}^{2}\right)\left(\frac{1}{1+{x}^{2}}\right)+2x\mathrm{arctan}x$$\left(1 + {x}^{2}\right) \left(\frac{1}{1 + {x}^{2}}\right) + 2 x \arctan x$
$1+2x\mathrm{arctan}x$$1 + 2 x \arctan x$

exponent of e

Finding the derivative of $y={e}^{x}$$y = {e}^{x}$ in first principles:

$\frac{d}{dx}{e}^{x}$$\frac{d}{\mathrm{dx}} {e}^{x}$

$=\underset{\delta \to 0}{lim}\left[{e}^{x+\delta }$$-{e}^{x}\right]/\delta$

$=\underset{\delta \to 0}{lim}\left[{e}^{x}×{e}^{\delta }$$-{e}^{x}\right]/\delta$

$=\underset{\delta \to 0}{lim}{e}^{x}\left[{e}^{\delta }$$-1\right]/\delta$

applying the standard limit $\underset{p\to 0}{lim}\left({e}^{p}-1\right)/p=1$${\lim}_{p \to 0} \left({e}^{p} - 1\right) / p = 1$

$={e}^{x}$$= {e}^{x}$

The above proves
$\frac{d}{dx}{e}^{x}={e}^{x}$$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

logarithm base e

Finding the derivative of $y=\mathrm{ln}x$$y = \ln x$ :

$y=\mathrm{ln}x$$y = \ln x$

${e}^{y}=x$${e}^{y} = x$

differentiating the equation
$\frac{d}{dx}{e}^{y}=1$$\frac{d}{\mathrm{dx}} {e}^{y} = 1$

applying chain rule $\frac{d}{dy}{e}^{y}\frac{dy}{dx}=1$$\frac{d}{\mathrm{dy}} {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

${e}^{y}\frac{dy}{dx}=1$${e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$x\frac{dy}{dx}=1$$x \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{dy}{dx}=\frac{1}{x}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x}$

The above proves
$\frac{d}{dx}\mathrm{ln}x=\frac{1}{x}$$\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$

exponent

Finding the derivative of $y={a}^{x}$$y = {a}^{x}$ :

substituting $a={e}^{\mathrm{ln}a}$$a = {e}^{\ln a}$

$y={e}^{x\mathrm{ln}a}$$y = {e}^{x \ln a}$

differentiating the equation
$\frac{dy}{dx}=\frac{d}{dx}{e}^{x\mathrm{ln}a}$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} {e}^{x \ln a}$

applying chain rule with $u=x\mathrm{ln}a$$u = x \ln a$
$\frac{dy}{dx}=\frac{d}{du}{e}^{u}\frac{d}{dx}\left(x\mathrm{ln}a\right)$$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{du}} {e}^{u} \frac{d}{\mathrm{dx}} \left(x \ln a\right)$

$\frac{dy}{dx}={e}^{u}×\mathrm{ln}a$$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \times \ln a$

$\frac{dy}{dx}={e}^{x\mathrm{ln}a}×\mathrm{ln}a$$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x \ln a} \times \ln a$

substituting ${e}^{\mathrm{ln}a}=a$${e}^{\ln a} = a$

$\frac{dy}{dx}={a}^{x}\mathrm{ln}a$$\frac{\mathrm{dy}}{\mathrm{dx}} = {a}^{x} \ln a$

The above proves
$\frac{d}{dx}{a}^{x}={a}^{x}\mathrm{ln}a$$\frac{d}{\mathrm{dx}} {a}^{x} = {a}^{x} \ln a$

summary

$\frac{d}{dx}{e}^{x}={e}^{x}$$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$
definition of $e$$e$ is rate of change is proportional to itself

$\frac{d}{dx}{a}^{x}={a}^{x}\mathrm{ln}a$$\frac{d}{\mathrm{dx}} {a}^{x} = {a}^{x} \ln a$
$a$$a$ equals ${e}^{\mathrm{ln}a}$${e}^{\ln a}$

$\frac{d}{dx}\mathrm{ln}x=\frac{1}{x}$$\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$
natural log is inverse of $e$$e$ power

example

What is the derivative of ${\mathrm{log}}_{10}x$${\log}_{10} x$?
Note: Use the identity ${\mathrm{log}}_{10}x=\frac{{\mathrm{log}}_{e}x}{{\mathrm{log}}_{e}10}$${\log}_{10} x = \frac{{\log}_{e} x}{{\log}_{e} 10}$

The answer is "$\frac{1}{x\mathrm{ln}10}$$\frac{1}{x \ln 10}$".
$\frac{d}{dx}{\mathrm{log}}_{10}x$$\frac{d}{\mathrm{dx}} {\log}_{10} x$
$=\frac{d}{dx}\frac{{\mathrm{log}}_{e}x}{{\mathrm{log}}_{e}10}$$= \frac{d}{\mathrm{dx}} \frac{{\log}_{e} x}{{\log}_{e} 10}$
$=\frac{1}{{\mathrm{log}}_{e}10}\frac{d}{dx}{\mathrm{log}}_{e}x$$= \frac{1}{{\log}_{e} 10} \frac{d}{\mathrm{dx}} {\log}_{e} x$
$=\frac{1}{{\mathrm{log}}_{e}10}×\frac{1}{x}$$= \frac{1}{{\log}_{e} 10} \times \frac{1}{x}$
$=\frac{1}{x{\mathrm{log}}_{e}10}$$= \frac{1}{x {\log}_{e} 10}$

What is the derivative of ${e}^{ax}$${e}^{a x}$?

The answer is "$a{e}^{ax}$$a {e}^{a x}$". Applying chain rule with $u=ax$$u = a x$,
$\frac{d}{du}{e}^{u}\frac{d}{dx}ax$$\frac{d}{\mathrm{du}} {e}^{u} \frac{d}{\mathrm{dx}} a x$
${e}^{u}×a$${e}^{u} \times a$
$a{e}^{ax}$$a {e}^{a x}$

Outline

The outline of material to learn "Differential Calculus" is as follows.

•   Detailed outline of Differential Calculus

→   Application Scenario

→   Differentiation in First Principles

→   Graphical Meaning of Differentiation

→   Differntiability

→   Algebra of Derivatives

→   Standard Results