what is the derivative of y=sin(cosx)?

Question

use the chain rule

Answer 1

y=sin(cosx)
dy/dx = cos(cosx)-sinx------------ use the chain rule
= -sincos(cosx)

Answer 2

This Site Might Help You.

RE:
what is the derivative of y=sin(cosx)?
use the chain rule

Answer 3

y=sin(cosx)
y' = cos(cosx) * -sinx

..you do the derivative of the outside then the derivitive of the inside.

Answer 4

y'=cos(cos(x))*-sin(x)
=-sin(x)cos(cos(x))

The chain rule says, first, take the derivative of the outer most function and leave the inner function as-is. Then take the derivative of the inner function to multiply the first.

Answer 5

Use the chain rule dy/dx = dy/du du/dx

Let u=cosx --> du/dx = -sinx

dy/dx = d[sin(u)]/dx
dy/dx = du/dx cos(u)
dy/dx = -sinx cos(u)
dy/dx = -sinx cos(cosx)

Answer 6

Chain rule states if f(x)=h(g(x))
then f'(x)=h'(g(x))g'(x)

In this case h(x)=sin(x) and h'(x)=cos(x)
and g(x)=cos(x) and g'(x)=-sin(x)

so f'(x)=h'(cos(x))(-sin(x))
=cos(cos(x))(-sin(x))
=-cos(cos(x))sin(x)