The derivative of $x^{\sin x}$ is, x>0
32 views
0 Votes
0 Votes

The derivative of $x^{\sin x}$ is, $x>0$

(A) $x^{\sin x-1} \sin x-x^{\sin x} \cos x \log x$

(B) $x^{\sin x-1} \sin x+x^{\sin x} \cos x \log x$

(C) $x^{\sin x-1} \sin x$

(D) None of the above

1 Answer

0 Votes
0 Votes
 
Best answer

SOLUTION —

Let $y=x^{\sin x}$

$\Rightarrow  \quad \log y  =\sin x \log x \\$

$\Rightarrow   \frac{1}{y} \frac{d y}{d x}  =\frac{\sin x}{x}+\log x \cos x \\$

$\Rightarrow  \frac{d y}{d x}  =x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right] \\$

$ =x^{\sin x-1} \sin x+x^{\sin x} \cos x \log x$

So, The correct option will be (B).

RELATED DOUBTS

1 Answer
0 Votes
0 Votes
28 Views
1 Answer
0 Votes
0 Votes
26 Views
1 Answer
0 Votes
0 Votes
39 Views
1 Answer
0 Votes
0 Votes
49 Views
1 Answer
0 Votes
0 Votes
27 Views
1 Answer
1 Vote
1 Vote
60 Views
1 Answer
0 Votes
0 Votes
30 Views
Peddia is an Online Question and Answer Website, That Helps You To Prepare India's All States Boards & Competitive Exams Like IIT-JEE, NEET, AIIMS, AIPMT, SSC, BANKING, BSEB, UP Board, RBSE, HPBOSE, MPBSE, CBSE & Other General Exams.
If You Have Any Query/Suggestion Regarding This Website or Post, Please Contact Us On : [email protected]

CATEGORIES