Question
Differentiate $\sin^{-1}\Big\{\frac{2^{\text{x}+1}\times3^\text{x}}{1+(36)^\text{x}}\Big\}$ with respect to x:
✓
Answer
We have $\text{y}=\sin^{-1}\Big\{\frac{2^{\text{x}+1}\times3^\text{x}}{1+(36)^\text{x}}\Big\}$
$\Rightarrow \text{y}=\sin^{-1}\Big\{\frac{2\times6^\text{x}}{1+6^{2\text{x}}}\Big\}$
Put $6^\text{x}=\tan\theta$
$\Rightarrow \theta=\tan^{-1}(6^\text{x})$
Now, $\text{y}=\sin^{-1}\Big\{\frac{2\tan\theta}{1+\tan^2\theta}\Big\}$
$\Rightarrow \text{y}=\sin^{-1}\big\{\sin2\theta\big\}$
$\Rightarrow \text{y}=2\theta$
$\Rightarrow \text{y}=2\tan^{-1}(6^\text{x})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\times\frac{1}{(6^\text{x})^2}\times6^\text{x}\log6$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2(\log6)6^\text{x}}{36^\text{x}}$
$\Rightarrow \text{y}=\sin^{-1}\Big\{\frac{2\times6^\text{x}}{1+6^{2\text{x}}}\Big\}$
Put $6^\text{x}=\tan\theta$
$\Rightarrow \theta=\tan^{-1}(6^\text{x})$
Now, $\text{y}=\sin^{-1}\Big\{\frac{2\tan\theta}{1+\tan^2\theta}\Big\}$
$\Rightarrow \text{y}=\sin^{-1}\big\{\sin2\theta\big\}$
$\Rightarrow \text{y}=2\theta$
$\Rightarrow \text{y}=2\tan^{-1}(6^\text{x})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\times\frac{1}{(6^\text{x})^2}\times6^\text{x}\log6$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2(\log6)6^\text{x}}{36^\text{x}}$
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