Question

Derivative

Answer 1

\[\LARGE \frac{x}{\sin^nx}\]

Answer 2

ATTEMPT:\[\LARGE \frac{\sin^n x-x^2 n \sin^{n-1}}{\sin^{2n}x^2}\]

Answer 3

@hartnn

Answer 4

how did u get x^2 ?
and do you know chain rule ?

Answer 5

\[\LARGE x(\frac{d(\sin^n x))}{dx}\]
and yes

Answer 6

using chain rule,
\((d/dx)\sin^n x = n \sin^{n-1}x (d/ dx[\sin x]) \)
got this ?

Answer 7

oh and btw its a NCERT que so I dont think chain rule will be used

Answer 8

without chain rule, this can't be solved..

Answer 9

\m/

Answer 10

okay thanks

Answer 11

also, in the quotient rule has square of denominator, and
\(\large (\sin^nx)^2= \sin^{2n}x\)
only.