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Question:

Differentiate $\sin ^{-1}\left\{\frac{2^{x+1} \cdot 3^{x}}{1+(36)^{x}}\right\}$ with respect to $x$.

Solution:

$y=\sin ^{-1}\left\{\frac{2^{x+1} \cdot 3^{x}}{1+(36)^{x}}\right\}$

$y=\sin ^{-1}\left\{\frac{2 \times 2^{x} \times 3^{x}}{1+\left(6^{2}\right)^{x}}\right\}$

$y=\sin ^{-1}\left\{\frac{2 \times 6^{x}}{1+\left(6^{x}\right)^{2}}\right\}$

Put $6^{x}=\tan \theta$

$y=\sin ^{-1}\left\{\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right\}$

Using $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

Now,

$y=\sin ^{-1}(\sin 2 \theta)$

$y=2 \theta$

$y=2 \tan ^{-1}\left(6^{x}\right)$

Differentiating w.r.t $x$, we get

$\frac{d y}{d x}=\frac{d}{d x}\left(2 \tan ^{-1} 6^{x}\right)$

$\frac{d y}{d x}=2 \times \frac{6^{x} \log 6}{1+\left(6^{x}\right)^{2}}$

$\frac{d y}{d x}=\frac{2 \times 6^{x} \log 6}{1+6^{2 x}}$

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