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If \( y = ( tan^{-1}x )^2\), show that \( (x^2 + 1 )^2 \: y_2 + 2x (x^2 +1 ) \: y_1 = 2 \)

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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:
We have $y=(\tan^{-1}x)^2$
$y=(\tan^{-1}x)^2$
$y_1=2\tan{-1}x\large\frac{d}{dx}$$(\tan^{-1}x)$
$y_1=2\tan^{-1}x.\large\frac{1}{1+x^2}$
$y_1=\large\frac{2\tan^{-1}x}{1+x^2}$
$(1+x^2)y_1=2\tan^{-1}x$
Step 2:
Differentiating with respect to $x$ we have
$(1+x^2)y_2+y_1(2x)=\large\frac{2}{1+x^2}$
$(1+x^2)^2y_2+2x(1+x^2)y_1=2$
Hence proved.
answered May 13, 2013 by sreemathi.v
 

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