Question

Differentiate with respect to x
1/sin^2x

Answer 1

Do you know how to rewrite exponents in the denominator like this: \[\LARGE \frac{1}{x^n}=x^{-n}\]

Answer 2

\[-\sin^2x/(\sin^2x)^2\]
\[-2cosxsinx/\sin^4x\]

Answer 3

or it will be better if u use Quotient rule

Answer 4

If i used quotient rule
\[1'*\sin^2x-(2cosxsinx)/\sin^4x\]
@Yahoo!

Answer 5

0*sin^2x-2cosxsinx/sin^4x

Answer 6

yeah...but that sin^ 4 x is for all terms since here the 1st term is 0 u dont need to consider it

Answer 7

okay thank you one more question if you don't mind

Answer 8

Sure..

Answer 9

\[\sin^2(2x)\] to find derivative do we use product rule?

Answer 10

u can.... but u can do it with product rule also think of the way u diff sin^2x in above case

Answer 11

there it was x here it is 2x

Answer 12

so we differentiate sin^2(2x)
sinx=cosx
2cosxsin2x

Answer 13

nope

differentiate sin^2(2x) = 2sin2x cos2x * 2 = 4 sin2x cos2x

Answer 14

f(x)=sin^2x
g(x)=2x
use chain rule: f'(g(x))*g'(x)'
2cos*sin*2x*2

Answer 15

e^(cos(x)^2) find derivative with respect to x