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If \(y=\sin ^{98}(x) \cdot \cos ^{39}(x)\), then find \(\frac{d y}{d x}\)
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Verified Answer
The correct answer is:
\(\left(98 \cos ^{99} x \cdot \sin ^{38} x\right)-\left(39 \sin ^{40} x \cdot \cos ^{97} x\right)\)
\(y=\sin ^{98} x \cdot \cos ^{39} x\)
\(\begin{aligned}
& \frac{d y}{d x}=\sin ^{98} x \cdot 39 \cos ^{38} x(-\sin x) \\
& +\cos ^{39 x} \cdot 98 \sin ^{97} x(\cos x) \\
& =98 \sin ^{97} x \cdot \cos ^{40} x-39 \sin ^{99} x \cdot \cos ^{38} x \\
& =98 \cos ^{99} x \cdot \sin ^{38} x-39 \sin ^{40} x \cdot \cos ^{97} x \\
\end{aligned}\)
\(\begin{aligned}
& \frac{d y}{d x}=\sin ^{98} x \cdot 39 \cos ^{38} x(-\sin x) \\
& +\cos ^{39 x} \cdot 98 \sin ^{97} x(\cos x) \\
& =98 \sin ^{97} x \cdot \cos ^{40} x-39 \sin ^{99} x \cdot \cos ^{38} x \\
& =98 \cos ^{99} x \cdot \sin ^{38} x-39 \sin ^{40} x \cdot \cos ^{97} x \\
\end{aligned}\)
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