Find the derivatives of f(g(x)) and g(f(x))

f(u)=u^3, g(x)= 1/1+x

Question

Find the derivatives of f(g(x)) and g(f(x))

f(u)=u^3,  g(x)= 1/1+x

anonymous · Answer

f= 3u^2  g= -1/(1-X)^2 i would think

mertsj · Answer

Satellite appears and the crowd gathers.

anonymous · Answer

just spacing out

anonymous · Answer

So you just find the derivatives and you don't have to use the chain rule?

anonymous · Answer

nvm idk dont you mean x instead of u for the f(u)= U^3

anonymous · Answer

$$f(g(x))=f(\frac{1}{1+x})=(\frac{1}{1+x})^3$$
$$f(g(x))'=3(\frac{1}{1+x})^2\times \frac{-1}{(1+x)^2}$$

anonymous · Answer

yes by the chain rule

anonymous · Answer

yeah satellite is correct i was simply using wut u game me w/o subbing in for the f(g(x)) part

anonymous · Answer

$$g(f(x))=g(x^3)=\frac{1}{1+x^3}$$
$$g(f(x))'=-\frac{1}{(1+x^3)^2}\times 3x$$

raspberryjam · Answer

Essentially you are finding the derivative of $$f(g(x))=(\frac{1}{1+x})^3$$ and $$g(f(x))=(\frac{1}{1+x^3})$$

anonymous · Answer

both clean up a tiny bit with some algebra

anonymous · Answer

Ah makes sense. I see where I went wrong. Thanks everyone.

anonymous · Answer

hey satellite are u good with calc 2? i got a few questions