If $y=\sin ^{98}(x) \cdot \cos ^{39}(x)$, then find $\frac{d y}{d x}$

Question

Aicharya · Accepted Answer

$y=\sin ^{98} x \cdot \cos ^{39} x$ $\begin{aligned} & \frac{d y}{d x}=\sin ^{98} x \cdot 39 \cos ^{38} x(-\sin x) \ & +\cos ^{39 x} \cdot 98 \sin ^{97} x(\cos x) \ & =98 \sin ^{97} x \cdot \cos ^{40} x-39 \sin ^{99} x \cdot \cos ^{38} x \ & =98 \cos ^{99} x \cdot \sin ^{38} x-39 \sin ^{40} x \cdot \cos ^{97} x \ \end{aligned}$

If \(y=\sin ^{98}(x) \cdot \cos ^{39}(x)\), then find \(\frac{d y}{d x}\)

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