Differentiation Question 421
Question: $ \frac{d}{dx}[{{\sin }^{n}}x\cos nx]= $
Options:
A) $ n{{\sin }^{n-1}}x\cos (n+1)x $
B) $ n{{\sin }^{n-1}}x\cos nx $
C) $ n{{\sin }^{n-1}}x\cos (n-1)x $
D) $ n{{\sin }^{n-1}}x\sin (n+1)x $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{d}{dx}[{{\sin }^{n}}x\cos nx]=n{{\sin }^{n-1}}x\cos x\cos nx-n\sin nx{{\sin }^{n}}x $
$ =n{{\sin }^{n-1}}x[\cos x\cos nx-\sin nx\sin x]=n{{\sin }^{n-1}}x\cos (n+1)x $ .
BETA
