question

[Nathan23]

Use the product rule to find the derivative.

f(x) = (3x^4 + 8)^2, I am getting f'(x) = 9x^16 + 64 am I messing this up?

 

[soroban]

Hello, Nathan23!

Use the product rule to find the derivative: $\displaystyle f(x)\:=\: (3x^4\,+\,8)^2$
I am getting $\displaystyle f'(x)\:=\:9x^{16}\,+\, 64\;\;$ Am I messing this up? $\displaystyle \;$ . . . you sure are!

First of all, they don't want you to multiply it out . . . they want you to use the Product Rule.

Secondly, you can't even multiply it out correctly! $\displaystyle \;\;(3x^4\,+\,8)(3x^4\,+\,8)\;=\;9x^8\,+\,48x\,+\,64$

 

[Nathan23]

Soroban I just did it again and got this.

f'(x) = 6x^4 + 16

 

[Unco]

No. Multi-choice?


Take the time to actually read Soroban's post.

 

[rommel902]

so
they want you to find the derivative of f(x)= (3x^4 +8)(3x^4 +8)
so f'(x)= (3x^4 + 8)(12x^3) + (3x^4 + 8)(12x^3)
so f'(x)= (36x^7 + 96x^3) + (36x^7 + 96x^3)
so f'(x)= 72x^7 + 192x^3 so f'(x)=24(3x^7 + 8x^3)

that is product rule which is if
f(x) = g(x)h(x)
then
f'(x) = g(x)h'(x) + h(x)g'(x)

an easier way is called the chain rule which for this problem is
f(x)= (3x^4 + 8)^2
f'(x)= 2(3x^4 + 8)(12x^3)
f'(x)= 2(36x^7 + 96x^3) or
f'(x)=24(3x^7 + 8x^3)

Does that explain it?

Peace,
KC