# If $y=\tan^{-1}x$, find $\large {\frac{d^2y}{dx^2}}$ in terms of y alone.

$\begin{array}{1 1}(A)\;-2\sin y.\cos^3y\\(B)\;4\sin y.\cos^3y\\(C)\;2\sin y.-\cos^3y\\(D)\;2\sin y.\cos^3y\end{array}$

Toolbox:
• To find $\large\frac{dy}{dx}$ for implict function $\large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}$$\phi(y).\large\frac{dy}{dx}$
• A function $f(x,y)$ is said to be implicit if it is jumbled in such a way,that it is not possible to write $y$ exclusively as a function of $x$.
• If $f'(x)$ is differentiable,we may differentiated this again w.r.t $x$.Then the LHS becomes $\large\frac{d}{dx}\big(\large\frac{dy}{dx}\big)$,which is called the second order derivative of $y$ w.r.t $x$.
Step 1:
Given $y=\tan^{-1}x$-----(1)
$\Rightarrow x=\tan y$
Differentiating w.r.t $x$
$\large\frac{dy}{dx}=\frac{1}{1+x^2}$-----(2)
Again differentiating equ(2) w.r.t $x$ by applying quotient rule we get,
$\large\frac{(1+x^2).0-1.(2x)}{(1+x^2)^2}=\frac{-2x}{(1+x^2)^2}$
$\Rightarrow \large\frac{-2x}{(1+x^2)^2}$
Step 2:
Substituting for $x$ we get,
$\Rightarrow \large\frac{-2\tan y}{(1+\tan^2x)^2}$
But $1+\tan^2y=\sec^2y$
$\large\frac{-2\tan y}{(1+\tan^2x)^2}=\large\frac{-2\tan y}{(\sec^2y)^2}$
But $\sec y=\large\frac{1}{\cos y}$
$\tan y=\large\frac{\sin y}{\cos y}$
Therefore $\large\frac{d^2y}{dx^2}=\large\frac{-2\sin y/\cos x}{1/\cos^4y}$
$\Rightarrow \large\frac{-2\sin y}{\cos y}$$.\cos^2y=-2\sin y.\cos^3y$
edited Mar 24, 2014