If f(x)=operatorname{cosec}^{-1}left[frac{10} | vedclass

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(A) Given $f(x)=\operatorname{cosec}^{-1}\left[\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right]$.Using the property $\operatorname{co

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$2^x \log 2$

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(A) Given $f(x)=\operatorname{cosec}^{-1}\left[\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right]$.Using the property $\operatorname{cosec}^{-1}(u) = \sin^{-1}(1/u)$,we get:$f(x)=\sin ^{-1}\left[\frac{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}{10}\right]$.We can write this as $f(x)=\sin ^{-1}\left[\frac{6}{10} \sin \left(2^x\right)-\frac{8}{10} \cos \left(2^x\right)\right]$.Let $\cos \alpha = \frac{6}{10}$ and $\sin \alpha = \frac{8}{10}$.Then $f(x)=\sin ^{-1}\left[\sin \left(2^x\right) \cos \alpha - \cos \left(2^x\right) \sin \alpha\right]$.Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we have:$f(x)=\sin ^{-1}\left[\sin \left(2^x-\alpha\right)\right] = 2^x - \alpha$.Differentiating with respect to $x$:$f^{\prime}(x) = \frac{d}{dx}(2^x - \alpha) = 2^x \log 2 - 0 = 2^x \log 2$.

If $f(x)=\operatorname{cosec}^{-1}\left[\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right]$,then $f^{\prime}(x)$ is equal to:

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