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If \( y = \tan^{-1}x\) show that $(1+x^2)\large\frac{d^2y}{dx^2}$$+2x\large\frac{dy}{dx}$$=0$

1 Answer

Toolbox:
  • $y=f(x)$
  • $\large\frac{dy}{dx}$$=f'(x)$
  • $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
  • $\large\frac{d}{dx}$$(\tan^{-1}x)=\large\frac{1}{1+x^2}$
Step 1:
$y=\tan^{-1}x$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=\frac{1}{1+x^2}$
Step 2:
Differentiating again w.r.t $x$ we get,
$\large\frac{d^2y}{dx^2}=\frac{(1+x^2)(0)-1(2x)}{(1+x^2)^2}$
$\Rightarrow \large\frac{-2x}{(1+x^2)^2}$
Step 3:
$\therefore (1+x^2)\large\frac{d^2y}{dx^2}$$+2x.\large\frac{dy}{dx}$$=(1+x^2)\large\frac{-2x}{(1+x^2)^2}$$+2x.\large\frac{1}{1+x^2}$
$\Rightarrow \large\frac{-2x+2x}{(1+x^2)}$
$\Rightarrow 0$
Hence proved.
answered Sep 30, 2013 by sreemathi.v
 
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