Question

What is the derivative of g(x)=(2+(x^2+1)^4)^3?

Answer 1

\[g(x)=(2+(x^2+1)^4)^3\]
You COULD multiply it out and brute force it, but the easy way would be to use chain rule.
\[f(x) = g(h(x))\]
\[f'(x) = g'(h(x))*h'(x)\]

your g() is simply the \[(h(x))^3\]
That derivative would yield a \[3*(h(x))^2 * \frac{d(h(x))}{dx}\]
The h(x) would be the 2+(x^2+1)^4 which would be another chain rule.

Answer 2

So, the derivative would be 2((x^2+3)^5+x)(10x(x^2+3)^4+1)?

Answer 3

remember that \[f(x)=a*x^n\]
\[f'(x)=n*a*x^{n-1}\]
The power decreases by 1, not increase.

Answer 4

Wait, uhh...Im kind of confused. let me show you my intermediate steps f'(x)=2((x^2+3)^5 +x) (5(x^2+3)^4+1)
=2((x^2+3)^5 +x) (5(x^2+3)^4+1)

Answer 5

Let me show you the first two steps.
I'll call \[x^2+1 = h(x)\]

\[g'(x) = 3*(2+h(x)^4)^{3-1}*\frac{d(2+h(x)^4)}{dx}\]
\[\frac{d(2+h(x)^4)}{dx}=\frac{d}{dx}2+\frac{d}{dx}h(x)^4=0+4*h(x)^{4-1}*h'(x)=4h(x)^3*h'(x)\]

Answer 6

Wow, sorry! When I was typing the wrong derivative for the question I asked. :( Haha.
but as it follows,
3(2+(x^2+1)^4)^2 (0+4(x^2+1)^3(2x) Right?...

Answer 7

yes that is correct. I was wondering where you were getting your expression from.

Answer 8

*I was typing the wrong derivative for the for question.
Anyways.
Then I would multiply 3 and 4 =12 then12 (2x), which would give me 24x
and 24x(x^2+1)^3((x^2+1)^4+2)^2?..

Answer 9

yes