If y=arctan x, show that dy/dx=1/1+x squared, using the inverse of trigonometric functions.
If y=arctan x, show that dy/dx=1/1+x squared, using the inverse of trigonometric functions.
I am not sure if this was the intended method - I always do these from scratch.
\(y=atan(x)\\ x=tan(y)\\ \frac{dx}{dy}=sec^2(y)\\ \frac{dy}{dx}=\frac{1}{sec^2(y)}\\ \frac{dy}{dx}=cos^2(y)\\~\\ Given\;\; tan(y)=\frac{x}{1}\;\;then\;\;\\ If \;\;opp=x\;\;and\;\;adj=1\;\;then\;\;hyp=\sqrt{1+x^2} \;\;so\\ cos^2(y)=(\frac{1}{\sqrt{1+x^2}})^2=\frac{1}{1+x^2}\\~\\ \frac{dy}{dx}=\frac{1}{1+x^2}\\\)