How do you differentiate #f(x)=(cos x+ 4sin^2x^2)^6# using the chain rule?

Question

William Abdalla · Accepted Answer

#6(cosx+4sin^2x^2)^5(-sinx+16xsinx^2cosx^2)#

In several stages.

First stage - chain rule for the bracket as a whole: #d/dx[(cosx+4sin^2x^2)^6]# #=# #6(cosx+4sin^2x^2)^5*d/dx[cosx+4sin^2x^2]#

Second stage - two terms, first one simple, second chain rule again: #d/dx[cosx]=-sinx# #d/dx[4sin^2x^2]=8sinx^2 *d/dx [sinx^2]#

Third stage - chain rule again #d/dx[sinx^2]=cosx^2*d/dx[x^2]#

Fourth stage - simple differentiation #d/dx[x^2]=2x#

Work back through the process: #d/dx[x^2]=2x# #d/dx[sinx^2]=2xcosx^2# #d/dx[4sin^2x^2]=16xsinx^2cosx^2#

#d/dx[(cosx+4sin^2x^2)^6]# #=# #6(cosx+4sin^2x^2)^5(-sinx+16xsinx^2cosx^2)#

Elijah Abram · Answer

undefined To differentiate $ f(x) = (\cos(x) + 4\sin^2(x^2))^6 $ using the chain rule, follow these steps:

1. Let $ u(x) = \cos(x) + 4\sin^2(x^2) $.
2. Let $ v(u) = u^6 $.
3. Apply the chain rule: $ \frac{dv}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} $.
4. Find $ \frac{du}{dx} $ by differentiating $ u(x) $ with respect to $ x $.
5. Find $ \frac{dv}{du} $ by differentiating $ v(u) $ with respect to $ u $.
6. Multiply $ \frac{dv}{du} $ and $ \frac{du}{dx} $ to get $ \frac{dv}{dx} $, which is the derivative of $ f(x) $.