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If $y=\cos \left(x^{\circ}\right), z=\cos x$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$\frac{\pi}{180} \sin \left(x^{\circ}\right) \operatorname{cosec} x$
We have,
$$
\begin{aligned}
y & =\cos x^0, z=\cos x \\
y & =\cos \frac{\pi x}{180} \\
\frac{d y}{d x} & =\frac{-\pi}{180} \sin \left(\frac{\pi x}{180}\right)=-\frac{\pi}{180} \sin x^{\circ} \\
\frac{d z}{d x} & =-\sin x \\
\frac{d y}{d z} & =\frac{\frac{d y}{d x}}{\frac{d z}{d x}}=\frac{-\frac{\pi}{180} \sin x^{\circ}}{-\sin x} \\
& =\frac{\pi}{180} \sin \left(x^{\circ}\right) \operatorname{cosec} x
\end{aligned}
$$
$$
\begin{aligned}
y & =\cos x^0, z=\cos x \\
y & =\cos \frac{\pi x}{180} \\
\frac{d y}{d x} & =\frac{-\pi}{180} \sin \left(\frac{\pi x}{180}\right)=-\frac{\pi}{180} \sin x^{\circ} \\
\frac{d z}{d x} & =-\sin x \\
\frac{d y}{d z} & =\frac{\frac{d y}{d x}}{\frac{d z}{d x}}=\frac{-\frac{\pi}{180} \sin x^{\circ}}{-\sin x} \\
& =\frac{\pi}{180} \sin \left(x^{\circ}\right) \operatorname{cosec} x
\end{aligned}
$$
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