Derivative of Trigonometric functions
Here will will discuss Derivative of sinx, cosx, tanx, cosecx, secx and cotx functions.
Derivative of sinx function
[katex]\frac d{dx}\left(\sin x\right)=\cos x[/katex]
[katex]y=\sin x[/katex]
[katex]y+\delta y=\sin\left(x+\delta x\right)[/katex]
[katex]y+\delta y-y=\sin\left(x+\delta x\right)-\sin x[/katex]
[katex]\boxed{\sin P-\sin Q=2\cos\left(\frac{P+Q}2\right)\sin\left(\frac{P-Q}2\right)}[/katex]
[katex]\delta y=2\cos\left(\frac{x+\delta x+x}2\right)\sin\left(\frac{x+\delta x-x}2\right)[/katex]
[katex]\delta y=2\cos\left(\frac{2x+\delta x}2\right)\sin\left(\frac{\delta x}2\right)[/katex]
dividing [katex]\delta x[/katex] on both sides.
[katex]\frac{\delta y}{\delta x}=2\frac{\cos\left(\frac{2x+\delta x}2\right)\sin\left(\frac{\delta x}2\right)}{\delta x}[/katex]
[katex]\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][/katex]
[katex]\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}}[/katex]
[katex]\frac{dy}{dx}=\cos\left(\frac{\displaystyle2x+0}{\displaystyle2}\right).1[/katex]
as we know [katex]\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}[/katex]
[katex]\boxed{\frac{dy}{dx}=\cos x}[/katex]
[katex]\boxed{\frac d{dx}\left(\sin x\right)=\cos x}[/katex]
This is the required derivative of sinx.
Derivative of cosx function
[katex]\frac d{dx}\left(\cos x\right)=-sin x[/katex]
[katex]y=\cos x[/katex]
[katex]y+\delta y=\cos\left(x+\delta x\right)[/katex]
[katex]y+\delta y-y=\cos\left(x+\delta x\right)-\cos x[/katex]
[katex]\boxed{\cos P-\cos Q=-2\sin\left(\frac{P+Q}2\right)\sin\left(\frac{P-Q}2\right)}[/katex]
[katex]\delta y=-2\sin\left(\frac{x+\delta x+x}2\right)\sin\left(\frac{x+\delta x-x}2\right)[/katex]
[katex]\delta y=-2\sin\left(\frac{2x+\delta x}2\right)\sin\left(\frac{\delta x}2\right)[/katex]
dividing [katex]\delta x[/katex] on both sides.
[katex]\frac{\delta y}{\delta x}=-2\frac{sin\left(\frac{2x+\delta x}2\right)\sin\left(\frac{\delta x}2\right)}{\delta x}[/katex]
[katex]\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][/katex]
[katex]\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}}[/katex]
[katex]\frac{dy}{dx}=-sin\left(\frac{\displaystyle2x+0}{\displaystyle2}\right).1[/katex]
As we know [katex]\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}[/katex]
[katex]\boxed{\frac{dy}{dx}=-sin x}[/katex]
[katex]\boxed{\frac d{dx}\left(\cos x\right)=-sin x}[/katex]
This is the required derivative of cosx.
Derivative of Tanx function
[katex]\frac d{dx}\left(Tan x\right)=sec^2 x[/katex]
[katex]y=Tan x[/katex]
[katex]y+\delta y=Tan\left(x+\delta x\right)[/katex]
[katex]y+\delta y-y=Tan\left(x+\delta x\right)-Tan x[/katex]
[katex]\delta y=\frac{\sin\left(x+\delta x\right)}{\cos\left(x+\delta x\right)}-\frac{\sin x}{\cos x}[/katex]
[katex]\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}[/katex]
using formula:
[katex]\boxed{\sin\alpha.\cos\beta-\cos\alpha.\sin\beta=\sin(\alpha-\beta)}[/katex]
[katex]\delta y=\frac{\sin\left(x+\delta x+x\right)}{\cos\left(x+\delta x\right).\cos x}[/katex]
[katex]\delta y=\frac{\sin\left(\delta x\right)}{\cos\left(x+\delta x\right).\cos x}[/katex]
dividing [katex]\delta x[/katex] on both sides.
[katex]\frac{\delta y}{\delta x}=\frac{\sin\left(\delta x\right)}{\cos\left(x+\delta x\right).\cos x.\delta x}[/katex]
Taking [katex]\underset{\delta x\rightarrow0}{lim}[/katex] on both sides.
[katex]\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}[/katex]
[katex]\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)}[/katex]
[katex]\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=\frac1{\cos\left(x\right).\cos x.}.1[/katex]
As we know [katex]\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}[/katex]
[katex]\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=\frac1{\cos^2x}[/katex]
[katex]\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=sec^2x[/katex]
This is the required derivative of tanx.
Derivative of Cosecx function
[katex]\frac d{dx}\left(cosec x\right)=-cosec x.cot x[/katex]
[katex]y=cosec x[/katex]
[katex]y+\delta y=cosec\left(x+\delta x\right)[/katex]
[katex]y+\delta y-y=cosec\left(x+\delta x\right)-cosec x[/katex]
[katex]\delta y=\frac{1}{\sin\left(x+\delta x\right)}-\frac{1}{\sin x}[/katex]
[katex]\delta y=\frac{sinx-\sin\left(x+\delta x\right)}{\sin\left(x+\delta x\right).\mathrm{sin}x}[/katex]
using formula
[katex]\boxed{\sin P-\sin Q=2\cos\left(\frac{P+Q}2\right)\sin\left(\frac{P-Q}2\right)}[/katex]
[katex]\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}[/katex]
dividing [katex]\delta x[/katex] on both sides.
[katex]\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}[/katex]
[katex]\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}}[/katex]
Taking [katex]\underset{\delta x\rightarrow0}{lim}[/katex] on both sides.
[katex]\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][/katex]
[katex]\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][/katex]
now using formula [katex]\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}[/katex]
[katex]\frac{dy}{dx}=-\left[\frac{\cos\left({\displaystyle\frac{2x+0}2}\right)}{\sin\left(x+0\right).\sin x.}\right].1[/katex]
[katex]\frac{dy}{dx}=-\left[\frac{\cos x}{\sin x.\sin x}\right].1[/katex]
[katex]\frac{dy}{dx}=-\frac{\cos x}{\sin x}.\frac1{\sin x}[/katex]
[katex]\boxed{\frac{dy}{dx}=-cosecx.cotx}[/katex]
This is the required derivative of cosecx.
Derivative of secx function
[katex]\frac d{dx}\left(sec x\right)=secx.tanx[/katex]
[katex]y=sec x[/katex]
[katex]y+\delta y=sec\left(x+\delta x\right)[/katex]
[katex]y+\delta y-y=sec\left(x+\delta x\right)-sec x[/katex]
[katex]\delta y=\frac{1}{\cos\left(x+\delta x\right)}-\frac{1}{\cos x}[/katex]
[katex]\delta y=\frac{cosx-\cos\left(x+\delta x\right)}{\cos\left(x+\delta x\right).\mathrm{cos}x}[/katex]
using formula
[katex]\boxed{\cos P-\sin Q=-2\sin\left(\frac{P+Q}2\right)\sin\left(\frac{P-Q}2\right)}[/katex]
[katex]\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}[/katex]
dividing [katex]\delta x[/katex] on both sides.
[katex]\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}[/katex]
[katex]\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}}[/katex]
Taking [katex]\underset{\delta x\rightarrow0}{lim}[/katex] on both sides.
[katex]\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][/katex]
[katex]\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][/katex]
now using formula [katex]\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}[/katex]
[katex]\frac{dy}{dx}=-\left[\frac{\cos\left({\displaystyle\frac{2x+0}2}\right)}{\sin\left(x+0\right).\sin x.}\right].1[/katex]
[katex]\frac{dy}{dx}=-\left[\frac{\cos x}{\sin x.\sin x}\right].1[/katex]
[katex]\frac{dy}{dx}=-\frac{\cos x}{\sin x}.\frac1{\sin x}[/katex]
[katex]\boxed{\frac{dy}{dx}=-cosecx.cotx}[/katex]
This is the required derivative of secx.
Derivative of Cotx function
[katex]\frac d{dx}\left(cot x\right)=-cosec^2 x[/katex]
[katex]y=cot x[/katex]
[katex]y+\delta y=cot\left(x+\delta x\right)[/katex]
[katex]y+\delta y-y=cot\left(x+\delta x\right)-cot x[/katex]
[katex]\delta y=\frac{\cos\left(x+\delta x\right)}{\sin\left(x+\delta x\right)}-\frac{\cos x}{\sin x}[/katex]
[katex]\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}[/katex]
using formula:
[katex]\boxed{\sin\alpha.\cos\beta-\cos\alpha.\sin\beta=\sin(\alpha-\beta)}[/katex]
[katex]\delta y=\frac{\sin\left(x-\delta x+x\right)}{\sin\left(x+\delta x\right).\sin x}[/katex]
[katex]\delta y=\frac{\sin\left(-\delta x\right)}{\sin\left(x+\delta x\right).\sin x}[/katex]
dividing [katex]\delta x[/katex] on both sides.
[katex]\frac{\delta y}{\delta x}=-\frac{\sin\left(\delta x\right)}{\sin\left(x+\delta x\right).\sin.\delta x}[/katex]
Taking [katex]\underset{\delta x\rightarrow0}{lim}[/katex] on both sides.
[katex]\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}[/katex]
[katex]\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)}[/katex]
[katex]\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-\frac1{\sin\left(x\right).\sin}.1[/katex]
As we know [katex]\boxed{\underset{\theta\rightarrow0}{lim}\left(\frac{\sin\theta}\theta\right)=1}[/katex]
[katex]\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-\frac1{\sin^2x}[/katex]
[katex]\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=-cosec^2x[/katex]
This is required derivative of cotx. (derivative of trigonometric functions)