Derivative of trigonometric function

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Derivative of trigonometric function :

$\frac{\text{d}}{\text{d}y}(sin x)= cos x$

$\frac{\text{d}}{\text{d}y}(cos x)= -sin x x$

$\frac{\text{d}}{\text{d}y}(tan x)= sec^{2} x$

You have already learnt the limits of trigonometric functions.

$\lim_{x \rightarrow 0}\frac{sin x}{x}=1$

$\lim_{x \rightarrow 0}\frac{1-cos x}{x}= 0$

Prove that: $\frac{\text{d}}{\text{d}x}(sin x)= cos x$

Proof : According to the definition of derivative
$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{f(x +\triangle x)-f(x)}{\triangle x}$

$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{sin(x +\triangle x)-sin(x)}{\triangle x}$

= $\lim_{\triangle x\rightarrow 0}\frac{[sin(x)cos(\triangle x)+cos(x)sin(\triangle x)]-[sin(x)]}{\triangle x}$

=$\lim_{\triangle x\rightarrow 0}\frac{sin(x)cos(\triangle x)-sin(x) +cos(x)sin(\triangle x)}{\triangle x}$

= $\lim_{\triangle x\rightarrow 0}\frac{sin(x)cos(\triangle x)-sin(x) +cos(x)sin(\triangle x)}{\triangle x}$

= $\lim_{\triangle x\rightarrow 0}\frac{sin(x)[cos(\triangle x)-1] +cos(x)sin(\triangle x)}{\triangle x}$

=$\lim_{\triangle x\rightarrow 0}\frac{cos(x)sin(\triangle x) -sin(x)[cos(\triangle x)-1]} {\triangle x}$

=$ \lim_{\triangle x\rightarrow 0}\frac{cos(x)sin(\triangle x)}{\triangle x} -\frac{sin(x)[cos(\triangle x)-1]} {\triangle x}$

=$ \lim_{\triangle x\rightarrow 0}\frac{cos(x)sin(\triangle x)}{\triangle x} -\lim_{\triangle x \rightarrow 0}\frac{sin(x)[cos(\triangle x)-1]} {\triangle x}$

= $ cos(x)\lim_{\triangle x\rightarrow 0}\frac{sin(\triangle x)}{\triangle x} -sin(x)\lim_{\triangle x \rightarrow 0}\frac{[cos(\triangle x)-1]} {\triangle x}$

= cos(x).1 - 0

$\frac{\text{d}}{\text{d}x}(sin (x))= cos x$

Prove that: $\frac{\text{d}}{\text{d}x}(cos(x))= -sin(x) x$

Proof : According to the definition of derivative
$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{f(x +\triangle x)-f(x)}{\triangle x}$

$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{cos(x +\triangle x)-cos(x)}{\triangle x}$

= $\lim_{\triangle x\rightarrow 0}\frac{[cos(x)cos(\triangle x)-sin(x)sin(\triangle x)]-[cos(x)]}{\triangle x}$

= $\lim_{\triangle x\rightarrow 0}\frac{cos(x)cos(\triangle x)-cos(x) - sin(x)sin(\triangle x)}{\triangle x}$

=$\lim_{\triangle x\rightarrow 0}\frac{cos(x)[cos(\triangle x)-1] -sin(x)sin(\triangle x)}{\triangle x}$

=$ \lim_{\triangle x\rightarrow 0}\frac{cos(x)[cos(\triangle x)-1]}{\triangle x} -\frac{sin(x)[sin(\triangle x)} {\triangle x}$

=$ \lim_{\triangle x\rightarrow 0}\frac{cos(x)[cos(\triangle x)-1]}{\triangle x} -\lim_{\triangle x \rightarrow 0}\frac{sin(x)sin(\triangle x)} {\triangle x}$

= 0 - sin(x).1

= -sin(x)

$\frac{\text{d}}{\text{d}x}(cos (x))= -sin (x)$

Prove that : $ \frac{\text{d}}{\text{d}x}(tan(x))= sec^{2}x$

Proof : $\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{f(x +\triangle x)-f(x)}{\triangle x}$

$\frac{\text{d}y}{\text{d}x}=\lim_{\triangle x\rightarrow 0}\frac{tan(x +\triangle x)-tan(x)}{\triangle x}$

=$\lim_{\triangle x\rightarrow 0}\frac{\frac{tan(x)+tan(\triangle x)}{1-tan(x)tan(\triangle x)}-tan(x)}{\triangle x}$

=$\lim_{\triangle x\rightarrow 0}\frac{\frac{tan(x) +tan(\triangle x)-tan(x)(1-tan(x)tan(\triangle x)}{(1-tan(x)tan(\triangle x)}}{\triangle x}$

=$\lim_{\triangle x\rightarrow 0}\frac{\frac{tan(x) tan(\triangle x)-tan(x)+tan^{2}(x)tan(\triangle x)}{(1-tan(x)tan(\triangle x)}}{\triangle x}$

=$\lim_{\triangle x\rightarrow 0}\frac{{ tan(\triangle x)(1+tan^{2}(x)}{}}{(1-tan(x)tan(\triangle x))\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{tan(\triangle x)}{\triangle x}\lim_{\triangle x \rightarrow 0}\frac{sec^{2}(x)}{(1-tan(x)tan(\triangle x))}$

=$1.\frac{sec^{2}(x)}{(1-tan(x)tan(0)}$

=$1 \frac{sec^{2}(x)}{(1-tan(x).0}$

$\frac{\text{d}}{\text{d}x}(tan(x))= sec^{2}x$


Examples on derivative of trigonometric function

Find the derivative of the following:
1) f(x) = 2 sin(x)
Solution:
As we know that $\frac{\text{d}}{\text{d}x}(sin(x))= cos(x)$

$\frac{\text{d}}{\text{d}x}(2 sin(x))= 2 cos(x)$

2) f(x) = $x^{2} $ + cos(x)

Solution : $\frac{\text{d}}{\text{d}x}(x^{2}) = 2x$ and $\frac{\text{d}}{\text{d}x}cos(x) = -sin(x)$

So, $\frac{\text{d}}{\text{d}x}(x^{2} $ + cos(x)) = 2x - cos(x)

3) f(x) = cos(x) - 4sin(x)

Solution : $\frac{\text{d}}{\text{d}x}(cos(x)) = -sin(x) $ and $\frac{\text{d}}{\text{d}x}sin(x) = cos(x)$

So, $\frac{\text{d}}{\text{d}x}(cos(x) - 4sin(x)) = -sin(x) - 4cos(x)$

12th grade math

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