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Solve any one of the following.
Differentiate sin⁻¹((2ˣ⁺¹· 3ˣ)/(1 + (36)ˣ)) with respect to x.
Differentiate $\sin^{-1}\left(\frac{2^{x+1}\cdot 3^x}{1 + (36)^x}\right)$ with respect to $x$.
Answer
Answer
$2^{x+1}\cdot3^x=2\cdot6^x$ and $1+36^x=1+(6^x)^2$. Put $6^x=\tan\theta$.
$\sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)=\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}(6^x)$ (for $x\le0$).
$\frac{d}{dx}=2\cdot\frac{6^x\log6}{1+36^x}$.
Final answer: $\frac{2\cdot 6^x\log 6}{1+36^x}$