Derivative of Trigonometric functions

^{Derivative of Trigonometric functions}

Here will will discuss Derivative of sinx, cosx, tanx, cosecx, secx and cotx functions.

Derivative of sinx function

$\frac d{dx}\left(\sin x\right)=\cos x$

$y=\sin x$

$y+\delta y=\sin\left(x+\delta x\right)$

$y+\delta y-y=\sin\left(x+\delta x\right)-\sin x$

$\boxed{\sin P-\sin Q=2\cos\left(\frac{P+Q}2\right)\sin\left(\frac{P-Q}2\right)}$

$\delta y=2\cos\left(\frac{x+\delta x+x}2\right)\sin\left(\frac{x+\delta x-x}2\right)$

$\delta y=2\cos\left(\frac{2x+\delta x}2\right)\sin\left(\frac{\delta x}2\right)$

dividing $\delta x$ on both sides.

$\frac{\delta y}{\delta x}=2\frac{\cos\left(\frac{2x+\delta x}2\right)\sin\left(\frac{\delta x}2\right)}{\delta x}$

$\underset{\delta x\rightarrow0}{lim}\frac{\displaystyle\delta y}{\displaystyle\delta x}=\underset{\delta x\rightarrow0}{lim}\left[2\frac{\displaystyle\cos\left(\frac{\displaystyle2x+\delta x}{\displaystyle2}\right)\sin\left(\frac{\displaystyle\delta x}{\displaystyle2}\right)}{\displaystyle\delta x}\right]$

$\frac{dy}{dx}=\underset{\delta x\rightarrow0}{lim}\cos\left(\frac{\displaystyle2x+\delta x}{\displaystyle2}\right).\underset{\delta x\rightarrow0}{lim}\frac{s\mathrm{in}\left(\frac{\textstyle\delta x}{\textstyle2}\right)}{\left(\frac{\textstyle1}{\textstyle2}\right){\displaystyle.}{\displaystyle\delta}{\displaystyle x}}$

$\frac{dy}{dx}=\cos\left(\frac{\displaystyle2x+0}{\displaystyle2}\right).1$

as we know $\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}$

$\boxed{\frac{dy}{dx}=\cos x}$

$\boxed{\frac d{dx}\left(\sin x\right)=\cos x}$

This is the required derivative of sinx.

Derivative of cosx function

$\frac d{dx}\left(\cos x\right)=-sin x$

$y=\cos x$

$y+\delta y=\cos\left(x+\delta x\right)$

$y+\delta y-y=\cos\left(x+\delta x\right)-\cos x$

$\boxed{\cos P-\cos Q=-2\sin\left(\frac{P+Q}2\right)\sin\left(\frac{P-Q}2\right)}$

$\delta y=-2\sin\left(\frac{x+\delta x+x}2\right)\sin\left(\frac{x+\delta x-x}2\right)$

$\delta y=-2\sin\left(\frac{2x+\delta x}2\right)\sin\left(\frac{\delta x}2\right)$

dividing $\delta x$ on both sides.

$\frac{\delta y}{\delta x}=-2\frac{sin\left(\frac{2x+\delta x}2\right)\sin\left(\frac{\delta x}2\right)}{\delta x}$

$\underset{\delta x\rightarrow0}{lim}\frac{\displaystyle\delta y}{\displaystyle\delta x}=\underset{\delta x\rightarrow0}{lim}\left[2\frac{\displaystyle-sin\left(\frac{\displaystyle2x+\delta x}{\displaystyle2}\right)\sin\left(\frac{\displaystyle\delta x}{\displaystyle2}\right)}{\displaystyle\delta x}\right]$

$\frac{dy}{dx}=\underset{\delta x\rightarrow0}{lim}-sin\left(\frac{\displaystyle2x+\delta x}{\displaystyle2}\right).\underset{\delta x\rightarrow0}{lim}\frac{s\mathrm{in}\left(\frac{\textstyle\delta x}{\textstyle2}\right)}{\left(\frac{\textstyle1}{\textstyle2}\right){\displaystyle.}{\displaystyle\delta}{\displaystyle x}}$

$\frac{dy}{dx}=-sin\left(\frac{\displaystyle2x+0}{\displaystyle2}\right).1$

As we know $\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}$

$\boxed{\frac{dy}{dx}=-sin x}$

$\boxed{\frac d{dx}\left(\cos x\right)=-sin x}$

This is the required derivative of cosx.

Derivative of Tanx function

$\frac d{dx}\left(Tan x\right)=sec^2 x$

$y=Tan x$

$y+\delta y=Tan\left(x+\delta x\right)$

$y+\delta y-y=Tan\left(x+\delta x\right)-Tan x$

$\delta y=\frac{\sin\left(x+\delta x\right)}{\cos\left(x+\delta x\right)}-\frac{\sin x}{\cos x}$

$\delta y=\frac{\sin\left(x+\delta x\right).\cos x-\cos\left(x+\delta x\right).\sin x}{\cos\left(x+\delta x\right).c\mathrm{cos}x}$

using formula:

$\boxed{\sin\alpha.\cos\beta-\cos\alpha.\sin\beta=\sin(\alpha-\beta)}$

$\delta y=\frac{\sin\left(x+\delta x+x\right)}{\cos\left(x+\delta x\right).\cos x}$

$\delta y=\frac{\sin\left(\delta x\right)}{\cos\left(x+\delta x\right).\cos x}$

dividing $\delta x$ on both sides.

$\frac{\delta y}{\delta x}=\frac{\sin\left(\delta x\right)}{\cos\left(x+\delta x\right).\cos x.\delta x}$

Taking $\underset{\delta x\rightarrow0}{lim}$ on both sides.

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=\underset{\delta x\rightarrow0}{lim}\frac{\sin\left(\delta x\right)}{\cos\left(x+\delta x\right).\cos x.\delta x}$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=\underset{\delta x\rightarrow0}{lim}\frac1{\cos\left(x+\delta x\right).\cos x.}.\underset{\delta x\rightarrow0}{lim}\frac{\sin\left(\delta x\right)}{\left(\delta x\right)}$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=\frac1{\cos\left(x\right).\cos x.}.1$

As we know $\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=\frac1{\cos^2x}$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=sec^2x$

This is the required derivative of tanx.

Derivative of Cosecx function

$\frac d{dx}\left(cosec x\right)=-cosec x.cot x$

$y=cosec x$

$y+\delta y=cosec\left(x+\delta x\right)$

$y+\delta y-y=cosec\left(x+\delta x\right)-cosec x$

$\delta y=\frac{1}{\sin\left(x+\delta x\right)}-\frac{1}{\sin x}$

$\delta y=\frac{sinx-\sin\left(x+\delta x\right)}{\sin\left(x+\delta x\right).\mathrm{sin}x}$

using formula

$\boxed{\sin P-\sin Q=2\cos\left(\frac{P+Q}2\right)\sin\left(\frac{P-Q}2\right)}$

$\delta y=\frac{2\cos\left({\displaystyle\frac{x+x+\delta x}2}\right).\sin\left(\frac{x-x-\delta x}2\right)}{\sin\left(x+\delta x\right).\sin x}$

dividing $\delta x$ on both sides.

$\frac{\delta y}{\delta x}=\frac{2\cos\left({\displaystyle\frac{2x+\delta x}2}\right).\sin\left(\frac{-\delta x}2\right)}{\sin\left(x+\delta x\right).\sin x.\delta x}$

$\frac{\delta y}{\delta x}=-\frac{\cos\left({\displaystyle\frac{2x+\delta x}2}\right).\sin\left(\frac{\delta x}2\right)}{\sin\left(x+\delta x\right).\sin x.\delta x.{\displaystyle\frac12}}$

Taking $\underset{\delta x\rightarrow0}{lim}$ on both sides.

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-\underset{\delta x\rightarrow0}{lim}\left[\frac{\cos\left({\displaystyle\frac{2x+\delta x}2}\right).\sin\left(\frac{\delta x}2\right)}{\sin\left(x+\delta x\right).\sin x.{\displaystyle\frac{\delta x}2}}\right]$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-\underset{\delta x\rightarrow0}{lim}\left[\frac{\cos\left({\displaystyle\frac{2x+\delta x}2}\right)}{\sin\left(x+\delta x\right).\sin x.}\right]\underset{\delta x\rightarrow0}{.lim}\left[\frac{\sin\left(\frac{\delta x}2\right)}{\frac{\delta x}2}\right]$

now using formula $\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}$

$\frac{dy}{dx}=-\left[\frac{\cos\left({\displaystyle\frac{2x+0}2}\right)}{\sin\left(x+0\right).\sin x.}\right].1$

$\frac{dy}{dx}=-\left[\frac{\cos x}{\sin x.\sin x}\right].1$

$\frac{dy}{dx}=-\frac{\cos x}{\sin x}.\frac1{\sin x}$

$\boxed{\frac{dy}{dx}=-cosecx.cotx}$

This is the required derivative of cosecx.

Derivative of secx function

$\frac d{dx}\left(sec x\right)=secx.tanx$

$y=sec x$

$y+\delta y=sec\left(x+\delta x\right)$

$y+\delta y-y=sec\left(x+\delta x\right)-sec x$

$\delta y=\frac{1}{\cos\left(x+\delta x\right)}-\frac{1}{\cos x}$

$\delta y=\frac{cosx-\cos\left(x+\delta x\right)}{\cos\left(x+\delta x\right).\mathrm{cos}x}$

using formula

$\boxed{\cos P-\sin Q=-2\sin\left(\frac{P+Q}2\right)\sin\left(\frac{P-Q}2\right)}$

$\delta y=-\frac{2\sin\left({\displaystyle\frac{x+x+\delta x}2}\right).\sin\left(\frac{x-x-\delta x}2\right)}{\cos\left(x+\delta x\right).\cos x}$

dividing $\delta x$ on both sides.

$\frac{\delta y}{\delta x}=-\frac{2\sin\left({\displaystyle\frac{2x+\delta x}2}\right).\sin\left(\frac{-\delta x}2\right)}{\cos\left(x+\delta x\right).\cos x.\delta x}$

$\frac{\delta y}{\delta x}=\frac{\sin\left({\displaystyle\frac{2x+\delta x}2}\right).\sin\left(\frac{\delta x}2\right)}{\cos\left(x+\delta x\right).\cos x.\delta x.{\displaystyle\frac12}}$

Taking $\underset{\delta x\rightarrow0}{lim}$ on both sides.

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=\underset{\delta x\rightarrow0}{lim}\left[\frac{\sin\left({\displaystyle\frac{2x+\delta x}2}\right).\sin\left(\frac{\delta x}2\right)}{\cos\left(x+\delta x\right).\cos x.{\displaystyle\frac{\delta x}2}}\right]$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-\underset{\delta x\rightarrow0}{lim}\left[\frac{\sin\left({\displaystyle\frac{2x+\delta x}2}\right)}{\cos\left(x+\delta x\right).\sin x.}\right]\underset{\delta x\rightarrow0}{.lim}\left[\frac{\cos\left(\frac{\delta x}2\right)}{\frac{\delta x}2}\right]$

now using formula $\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}$

$\frac{dy}{dx}=-\left[\frac{\cos\left({\displaystyle\frac{2x+0}2}\right)}{\sin\left(x+0\right).\sin x.}\right].1$

$\frac{dy}{dx}=-\left[\frac{\cos x}{\sin x.\sin x}\right].1$

$\frac{dy}{dx}=-\frac{\cos x}{\sin x}.\frac1{\sin x}$

$\boxed{\frac{dy}{dx}=-cosecx.cotx}$

This is the required derivative of secx.

Derivative of Cotx function

$\frac d{dx}\left(cot x\right)=-cosec^2 x$

$y=cot x$

$y+\delta y=cot\left(x+\delta x\right)$

$y+\delta y-y=cot\left(x+\delta x\right)-cot x$

$\delta y=\frac{\cos\left(x+\delta x\right)}{\sin\left(x+\delta x\right)}-\frac{\cos x}{\sin x}$

$\delta y=\frac{\cos\left(x+\delta x\right).\sin x-\sin\left(x+\delta x\right).\cos x}{\sin\left(x+\delta x\right).\mathrm{sin}x}$

using formula:

$\boxed{\sin\alpha.\cos\beta-\cos\alpha.\sin\beta=\sin(\alpha-\beta)}$

$\delta y=\frac{\sin\left(x-\delta x+x\right)}{\sin\left(x+\delta x\right).\sin x}$

$\delta y=\frac{\sin\left(-\delta x\right)}{\sin\left(x+\delta x\right).\sin x}$

dividing $\delta x$ on both sides.

$\frac{\delta y}{\delta x}=-\frac{\sin\left(\delta x\right)}{\sin\left(x+\delta x\right).\sin.\delta x}$

Taking $\underset{\delta x\rightarrow0}{lim}$ on both sides.

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-\underset{\delta x\rightarrow0}{lim}\frac{\sin\left(\delta x\right)}{\sin\left(x+\delta x\right).\sin x.\delta x}$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-\underset{\delta x\rightarrow0}{lim}\frac1{\sin\left(x+\delta x\right).sinx.}.\underset{\delta x\rightarrow0}{lim}\frac{\sin\left(\delta x\right)}{\left(\delta x\right)}$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-\frac1{\sin\left(x\right).\sin}.1$

As we know $\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-\frac1{\sin^2x}$

$\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-cosec^2x$

This is required derivative of cotx. (derivative of trigonometric functions)

Here will will discuss Derivative of sinx, cosx, tanx, cosecx, secx and cotx functions.

Derivative of sinx function

Derivative of cosx function

Derivative of Tanx function

Derivative of Cosecx function

Derivative of secx function

Derivative of Cotx function

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