Question

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

• A. $\left(98 \cos ^{99} x \cdot \sin ^{38} x\right)+\left(39 \sin ^{40} x \cdot \cos ^{97} x\right)$
• B. $\left(98 \cos ^{98} x \cdot \sin ^{39} x\right)-\left(40 \sin ^{39} x \cdot \cos ^{98} x\right)$
• C. $\left(98 \cos ^{99} x \cdot \sin ^{38} x\right)-\left(39 \sin ^{40} x \cdot \cos ^{97} x\right)$
• D. $\left(98 \cos ^{98} x \cdot \sin ^{39} x\right)+\left(39 \sin ^{40} x \cdot \cos ^{97} x\right)$

Answer 1

Answer: (C)

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