Differentiating Special Functions

Differentiating Special Functions – HMC Calculus Tutorial

In this tutorial, we review the differentiation of trigonometric, logarithmic, and exponential functions.

Trigonometric Functions

The derivatives of the basic trigonometric functions are given here for reference. \[\begin{array}{rr} f(x) & f'(x)\\ ~\\ \qquad\qquad\sin x & \cos x\\ \cos x & -\sin x\\ \tan x & \sec^2 x\\ \sec x & \sec x \tan x\\ \csc x & \qquad\qquad -\csc x\cot x\\ \cot x & -\csc^2 x\\ ~\\ \end{array}\] The derivatives of $\sin x$ and $\cos x$ can be derived using the limit definition of the derivative.

Limit Definition of the Derivative
Limit Definition of the Derivative

\[f'(x)=\lim_{\Delta x\to 0}\, \frac{f(x+\Delta x)-f(x)}{\Delta x}\]

For $\sin x$, \begin{eqnarray*} \frac{d}{dx}(\sin x)&=&\lim_{h\to 0} \frac{\sin (x+h)-\sin (x)}{h}\\ &=& \lim_{h\to 0} \frac{(\sin x\cos h+\cos x\sin h)-\sin x}{h}\\ &=& \lim_{h\to 0} \left[\sin x\frac{\cos h -1}{h}+\cos x\frac{\sin h}{h}\right]\\ &=&\sin x\lim_{h\to 0}\left[\frac{\cos h-1}{h}\right]+\cos x\lim_{h\to 0}\left[\frac{\sin h}{h}\right]\\ &=& \sin x (0)+\cos x (1)\\ &=& \cos x. \end{eqnarray*} The derivative of $\cos x$ is derived analogously. Then the remaining derivatives can be derived using the quotient rule, since all the other trigonometric functions are quotients involving $\sin x$ and $\cos x$.


The derivative of $\tan (x^2)$ is $\displaystyle \sec^2(x^2)\cdot\frac{d}{dx}(x^2) =2x\sec^2(x^2)$ by the chain rule.

Logarithmic Functions

By the definition of the natural logarithm, $\displaystyle \frac{d}{dx}[\ln x] =\frac{1}{x}$ for $x>0$. Also, $\displaystyle \frac{d}{dx} [\ln |x|]=\frac{1}{x}$ for all $x\neq 0$. To see this, suppose $x<0$. Then $\ln |x|=\ln (-x)$.

So \begin{eqnarray*} \frac{d}{dx}[\ln |x|]&=&\frac{d}{dx}[\ln (-x)]\\ &=& \frac{d}{dx}(-x)\left(\frac{1}{-x}\right)\\ &=& (-1)\left(\frac{1}{-x}\right)\\ &=& \frac{1}{x}. \end{eqnarray*}


By the chain rule, the derivative of $\ln (x^3+5)$ is $\displaystyle \frac{d(x^3+5)}{dx}\cdot\frac{1}{x^3+5}=\frac{3x^2}{x^3+5}$.

Exponential Functions

There is an elegant way to show that $\displaystyle \frac{d}{dx}\left[e^x\right]=e^x$. We start with the identity $\displaystyle \ln (e^x)=x$. Differentiating both sides, \begin{eqnarray*} \frac{d}{dx}[\ln (e^x)]&=&\frac{d}{dx}(x)\\ \frac{d}{dx}[\ln (e^x)]&=& 1\\ \frac{d}{dx}(e^x)\cdot \frac{1}{e^x}&=&1\\ \frac{d}{dx}(e^x)&=&e^x. \end{eqnarray*} Since $e^x$ is never $0$, this derivation holds for all $x$.


The derivative of $\displaystyle e^{-3x+2}$ is $\displaystyle e^{-3x+2}\cdot \frac{d}{dx}(-3x+2)=-3e^{-3x+2}$.

Key Concepts

\[\begin{array}{rr} f(x) & f'(x)\\ ~\\ \qquad\qquad\sin x & \cos x\\ \cos x & -\sin x\\ \tan x & \sec^2 x\\ \sec x & \sec x \tan x\\ \csc x & \qquad\qquad -\csc x\cot x\\ \cot x & -\csc^2 x\\ \ln x & \frac{1}{x}\\ e^x & e^x\\ ~\\ \end{array}\]

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