If y=e^{sin left(operatorname{cosec}^{-1} xri | vedclass

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(B) Given $y=e^{\sin \left(\operatorname{cosec}^{-1} x\right)}$.Since $\operatorname{cosec}^{-1} x = \sin^{-1} \left(\frac{1}{x}\right)$,we have $\sin

Vedclass · Accepted Answer

$-\frac{e^{\frac{1}{x}}}{x^{2}}$

Vedclass · Answer

(B) Given $y=e^{\sin \left(\operatorname{cosec}^{-1} x\right)}$.Since $\operatorname{cosec}^{-1} x = \sin^{-1} \left(\frac{1}{x}\right)$,we have $\sin \left(\operatorname{cosec}^{-1} x\right) = \sin \left(\sin^{-1} \frac{1}{x}\right) = \frac{1}{x}$.Therefore,the function simplifies to $y = e^{\frac{1}{x}}$.Now,differentiating with respect to $x$ using the chain rule:$\frac{d y}{d x} = \frac{d}{d x} \left(e^{\frac{1}{x}}\right) = e^{\frac{1}{x}} \cdot \frac{d}{d x} \left(\frac{1}{x}\right)$.Since $\frac{d}{d x} \left(\frac{1}{x}\right) = -\frac{1}{x^2}$,we get $\frac{d y}{d x} = e^{\frac{1}{x}} \left(-\frac{1}{x^2}\right) = -\frac{e^{\frac{1}{x}}}{x^2}$.

If $y=e^{\sin \left(\operatorname{cosec}^{-1} x\right)}$,then $\frac{d y}{d x}=$

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