Question

Please help me with this calculus problem!



Differentiate sin^2(cos^3x+tanx)

Answer 1

Any multiple choice option?

Answer 2

Nope.

Answer 3

Click this http://www.wolframalpha.com/input/?i=Differentiate+sin%5E2%28cos%5E3x%2Btanx%29

Answer 4

I already had that open :P, it just didnt seem right...

Answer 5

I think it is somewhat right.

Answer 6

So, I should follow that?

Answer 7

I guess so

Answer 8

sin^(2)(cos^3x+tanx)
use chain rule

g(x) = x^(2) g'(x) = 2x
s(x) = sin(cos^3x+tanx) s'(x) = (cos^3x+tanx)' = -3(cos^(2)(x))sin(x) + sec^(2)(x)

Now we have to differentiate both cos^(3)x and tan(x)
l(x) = x^(3) l'(x) = 3x^(2)
m(x) = cos(x) m'(x) = -sin(x)

o(x) = tan(x) o'(x) = sec^(2)(x)

Now we put it all together

f'(x) = 2sin(cos^3x+tanx) * -3((cos^(2)(x))sin(x) + sec^(2)(x))
f'(x) = -6sin(cos^3x+tanx)(cos^(2)(x))sin(x) + sec^(2)(x))

Answer 9

you can probably simplify it more but meh

Answer 10

Have you clicked the link?

Answer 11

note the derivative of cos^(3)x = -3(cos^(2)(x))sin(x)

Answer 12

I hope you can follow what I did to solve this

Answer 13

Yeah, dw about it

Answer 14

ok then I wont :\