How do you use the chain rule to differentiate #y=cos((4x)^3)#?

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Answer 1 · 2017-07-31T16:25:56

#y=cos((4x)^3) => y'=-192x^2sin((4x)^3)#

The Chain Rule States that

#(f(g(x))'=f'(g(x))(g'(x))#

in the case of #y=cos((4x)^3)#

#f(x)=cos(x)#

and

#g(x)=(4x)^3#

#f(x)=cos(x) => f'(x)=-sin(x)#

and #f'(g(x))=-sin((4x)^3)#

Also since #(ab)^x=a^x(b^x)#

then #(4x)^3=4^3(x^3)=64x^3#

And since #f'(cx)=c(f'(x))#

#g'(x)=(64x^3)=64(x^3)'#

and since the power rule states that #(x^n)'=nx^(n-1)#

#g'(x)=192x^2#

Then we plug in and get

#y'=-192x^2sin((4x)^3)#

Answer 2 · 2017-07-31T16:28:15

#d/(dx) [cos((4x)^3)] = color(blue)(-192x^2·sin(64x^3)#

We're asked to find the derivative

#d/(dx) [cos((4x)^3)]#

which is also

#d/(dx) [cos(64x^3)]#

Using the chain rule, which for this situation is

#d/(dx)[cos(64x^3)] = d/(du) [cosu] (du)/(dx)#

where

#= -sin(64x^3)·d/(dx)[64x^3]#

Using the power rule on the #64x^3# term:

#d/(dx) [64x^3] = 3·64x^(3-1)#

#= color(blue)(ul(-192x^2·sin(64x^3)#