Differentiate with respect to x
1/sin^2x

Question

Differentiate with respect to x
1/sin^2x

kainui · Answer

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

abmon98 · Answer

$$-\sin^2x/(\sin^2x)^2$$
$$-2cosxsinx/\sin^4x$$

anonymous · Answer

or it will be better if u use Quotient rule

abmon98 · Answer

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

abmon98 · Answer

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

anonymous · Answer

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

abmon98 · Answer

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

anonymous · Answer

Sure..

abmon98 · Answer

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

anonymous · Answer

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

anonymous · Answer

there it was x  here it is 2x

abmon98 · Answer

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

anonymous · Answer

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

abmon98 · Answer

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

abmon98 · Answer

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