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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Differentiate the following with respect to $x: \sin^{-1}\bigg(\large \frac{2^{x+1}.3^x}{1+(36)^x}\bigg)$

This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.
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  • $\large \frac{2tan\theta}{1+tan^2\theta}$$=sin2\theta$
Given $y= \sin^{-1}\bigg(\large \frac{2^{x+1}.3^x}{1+(36)^x}\bigg)$
$\Rightarrow y= \sin^{-1}\bigg(\large \frac{2^{x+1}.3^x}{1+(36)^x}\bigg)$
$\Rightarrow y =\sin^{-1}\bigg(\large \frac{2.6^x}{1+(6^x)^2}\bigg)$
Let $6^x=tan\theta$ $\rightarrow y=sin^{-1}\bigg(\large \frac{2tan\theta}{1+tan^2\theta}\bigg)$
$\large \frac{2tan\theta}{1+tan^2\theta}$$=sin2\theta$
$\Rightarrow y=sin^{-1}(sin2\theta) = 2\theta$
Since $6^x=tan\theta,\; y = 2tan^{-1}(6^x)$
Differentiating both sides, $\large \frac{dy}{dx}$$= 2.\large\frac{1}{1+(6^x)^2}$$\;6^xlog 6$
$\Rightarrow \large \frac{dy}{dx}=\frac{2log 6.6^x}{1+36^x}$
answered Mar 21, 2013 by sreemathi.v
edited Mar 22, 2013 by balaji.thirumalai
 

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