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

derivative of trigonometric functions
\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)

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Azhar Ali

Azhar Ali

I graduated in Mathematics from the University of Sargodha, having master degree in Mathematics.

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