The group of function that we differentiate are the trigonometric functions. Once we have a formula for the derivative of the sine and cosine function, formula for the rest of all the trigonometric function will follow from the quotient rule. That is we can derive the formula for the remaining four functions by rewriting these functions in terms of the sine and cosine functions. The derivative of $\sin x$ utilizes the following two limit results.

$\lim_{x \to 0}$$\frac{\sin x}{x}$ = $1$ and $\lim_{x\to 0}$$\frac{\cos x – 1}{x}$ = $0$

There is no general rule for determining the derivative of trigonometric function. Each trigonometric function has a unique derivative. The trigonometric functions and its derivatives are given below

$\frac{\mathrm{d} \sin x}{\mathrm{d} x}$ = $\cos x$

$\frac{\mathrm{d} \cos x}{\mathrm{d} x}$ = $-\sin x$

$\frac{\mathrm{d} \tan x}{\mathrm{d} x}$ = $\sec^{2} x$

$\frac{\mathrm{d} \cot x}{\mathrm{d} x}$ = $-\csc^{2} x$

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

$\frac{\mathrm{d} \csc x}{\mathrm{d} x}$ = $-\csc x\cot x$
Derivative of trigonometric function examples given below

Solved Examples

Question 1: Find $\frac{\mathrm{d} \sin x}{\mathrm{d} x}$ ?
Solution:
 
Let $f(x)$ = $\sin x$

$f(x+\delta x)$ = $\sin (x+\delta x)$

$\Rightarrow$ $\lim_{\delta x \to 0}$$\frac{f(x+\delta x)-f(x)}{\delta x}$ = $\lim_{\delta x \to 0}$$\frac{\sin (x+\delta x)-\sin (x)}{\delta x}$  

(We have $\sin (A)-\sin(B)$ = $2\sin(\frac{A-B}{2}).\cos(\frac{A+B}{2})$)

= $\lim_{\delta x \to 0}$$\frac{2\cos (\frac{x+\delta x+x}{2}).\sin (\frac{x+\delta x-x}{2})}{\delta x}$  

= $\lim_{\delta x \to 0}\cos (x+$$\frac{\delta x}{2}).\frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}}$

= $\cos x .1$ = $\cos x$

$\Rightarrow$ $\frac{\mathrm{d} f(x)}{\mathrm{d} x}$ = $\cos x$

Hence $\frac{\mathrm{d} \sin x}{\mathrm{d} x}$ = $\cos x$

 

Question 2: Find $\frac{\mathrm{d} \cos x}{\mathrm{d} x}$?
Solution:
 
Let $f(x)$ = $\cos x$

$f(x+\delta x)$ = $\cos (x+\delta x)$

$\Rightarrow$ $\lim_{\delta x \to 0}$$\frac{f(x+\delta x)-f(x)}{\delta x}$ = $\lim_{\delta x \to 0}$$\frac{\cos (x+\delta x)-\cos (x)}{\delta x}$  

(We have $\cos (A)-\cos(B)$ = $-2\sin($$\frac{A+B}{2}).\sin(\frac{A-B}{2})$)

= $\lim_{\delta x \to 0}$$\frac{-2\sin (\frac{x+\delta x+x}{2}).\sin (\frac{x+\delta x-x}{2})}{\delta x}$   

= $\lim_{\delta x \to 0}-\sin (x+$$\frac{\delta x}{2}).\frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}}$

= $-\sin x .1$ = $-\sin x$

$\Rightarrow$ $\frac{\mathrm{d} f(x)}{\mathrm{d} x}$ = $-\sin x$

Hence $\frac{\mathrm{d} \cos x}{\mathrm{d} x}$ = $-\sin x$


 

Question 3: Find $\frac{\mathrm{d} \tan x}{\mathrm{d} x}$?
Solution:
 
Let $f(x)$ = $\tan x$

we know that $\tan x$ = $\frac{\sin x}{\cos x}$

$\frac{\mathrm{d} \tan x}{\mathrm{d} x}$ = $\frac{\mathrm{d} \frac{\sin x}{\cos x}}{\mathrm{d} x}$

= $\frac{\cos x \cos x-\sin x.(-\sin x) }{(\cos x)^{2}}$

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

[but $(\cos x)^{2}+(\sin x)^{2}= 1$]

$\frac{\mathrm{d} \tan x}{\mathrm{d} x}$ = $\frac{1}{(\cos x)^{2}}$

= $\sec^{2}x$

 

Question 4: Perform the operation of differentiation for the function $y$ = $\cos(x^{3})$
Solution:
 
Let $u$ = $x^{3}$ and we can write $y$ as $y$ = $\cos u$

 By applying the formula $\frac{\mathrm{d} y}{\mathrm{d} x}$ = $\frac{\mathrm{d} y}{\mathrm{d} u}$ . $\frac{\mathrm{d} u}{\mathrm{d} x}$ 

= $\frac{\mathrm{d} \cos u}{\mathrm{d} u}$.$\frac{\mathrm{d} x^{3}}{\mathrm{d} x}$

= $(-\sin u).(3x^{2})$

= $(-\sin (x^{3})).(3x^{2})$

= $-3x^{2}\sin (x^{3})$

 

Question 5: Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ if $y$ = $x\sin x$
Solution:
 

Let $y$ = $x\sin x$

$\frac{\mathrm{d} y}{\mathrm{d} x}$ = $\frac{\mathrm{d} x\sin x}{\mathrm{d} x}$

= $x$ $\frac{\mathrm{d} \sin x}{\mathrm{d} x}$ + $\sin x$ $\frac{\mathrm{d} x}{\mathrm{d} x}$

= $x\cos x + \sin x$