# Calculus I problem I can't solve

Here's the problem:

$$d/dx(f(3x^5)) = 8x^2$$

Find $$f'(x)$$

After applying the chain rule:

$$f'(3x^5)(15x^4) = 8x^2$$

$$f'(3x^5) = 8/(15x^2)$$

It's not apparent to me how I proceed from here to find $$f'(x)$$. I tried dividing the expression on the right by three and taking the fifth root but that does not seem to be right. Any help would be appreciated!

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HallsofIvy
I see what you're saying, but isn't du/dx equal to $$15x^4$$, not $$15x$$? That would make df/du equal to $$(8/15x^2)$$. Therefore, substituting x= (u/3)1/5 gives a result of (8/15)(3^(2/5)/u^(2/5)).
Substituting u by x to get f'(x) would get a final solution of $$8/15x^2$$, which is what I started with above for $$f'(3x^5)$$, which I don't think is the correct answer.