If $y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^2}}{13}\right)$, then $\frac{d y}{d x}=$

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Aicharya · Accepted Answer

$\begin{aligned} & y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^2}}{13}ight) \ & \Rightarrow y=\sin -1\left(x \cdot \frac{5}{13}+\frac{12}{13} \sqrt{1-x^2}ight)\end{aligned}$ [let $x=\sin 	heta$ i.e., $	heta=\sin ^{-1} X$ and $\cos 	heta=\sqrt{1-X^2}$ also let $\frac{5}{13}=\cos \alpha$ i.e. $\left.\frac{12}{13}=\sin \alphaight]$ $\begin{aligned} & \Rightarrow y=\sin ^{-1}(\sin 	heta \cdot \cos \alpha+\cos 	heta \cdot \sin \alpha) \ & \Rightarrow y=\sin -1 \sin (	heta+\alpha) \ & \Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}+0=\frac{1}{\sqrt{1-x^2}}\end{aligned}$

If $y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^2}}{13}\right)$, then $\frac{d y}{d x}=$

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