Thread: f'(5x^4) of f(x)=x^5

1. f'(5x^4) of f(x)=x^5

Hi, I was wondering how to do these types of derivatives... I have several to do, but I would like to see only one of them solved so that I can do the others. The course is about several variable calculus...

The problem says:
"If f(x)=x^5, find

f'(5x^4), f(x^2), d/dx(f(x^2)), f'(x^2)"

2. Re: f'(5x^4) of f(x)=x^5

Originally Posted by juanma101285
The problem says:
"If f(x)=x^5, find
f'(5x^4), f(x^2), d/dx(f(x^2)), f'(x^2)"
If $\displaystyle f(x)=x^5$ then $\displaystyle f'(x)=5x^4$.

Thus $\displaystyle f'(5x^4)=5(5x^4)^4=5^5x^{16}.$

3. Re: f'(5x^4) of f(x)=x^5

This is just composition of functions. If I asked you to find f'(x), you'd probably have no problem doing that. Here's a different example:

f(x)=x^2+5

Find f'(x+3), f(x^5), d/dx(f(x^3)), f'(x^3)

First, let's calculate f'(x).
f'(x)=2x

Now, remember how to do composition of functions?
f'(x+3)=2(x+3)=2x+6

The second function is even easier, we just compose f with x^5:
f(x^5)=(x^5)^2+2 = x^10+2

Now we're back to derivatives, and some tricky notation. The book is trying to trip you up into thinking that, since d/dx(f(x))=f'(x), then d/dx(f(x^3))=f'(x^3). This is not the case. Remember the chain rule:

d/dx(f(g(x)) = f'(g(x))g'(x)

So, with g(x)=x^3, g'(x)=3x^2 and therefore
d/dx(f(x^3))=[f'(x^3)]g'(x)=[2(x^3)](3x^2)=6x^5

Finally, the last one is just there to possibly alert those people who tripped up on the third one that they should reconsider their answer. It's just simple composition, like before:

f'(x^2)=2(x^2)=2x^2