# How I can solve this derivative problem?

Suppose $$f'(x)$$ = 2$$x^4$$ $$f(x^2)$$ and $$f(1)$$ = 2.

Find $$f''(x)$$ = $${d}/{dx}$$ $$f'(x)$$

I'm having trouble starting this. I was wondering if I work backwards from 𝑓(1) = 2 OR use $$f(x^2)$$ somehow.

How can I solve this?

EDIT: So it looks like my friend miswrote the question, it's now edited. It should be $$f'(x)$$ = 2$$x^4$$ $$f(x^2)$$ NOT $$f(x)$$ = 2$$x^4$$ $$f(x^2)$$

• It's impossible to have $f(1)=2$ since $f(1) = 2\cdot 1^4 \cdot f(1) \iff f(1) = 0$. Jun 16 '20 at 19:59
• Yeah, this doesn't make sense. Where did you get this question from? Jun 16 '20 at 20:02
• @Savio How did you get 𝑓(1) = 0? Jun 16 '20 at 20:07
• @Riemann'sPointyNose The question was one my friend sent me from a tutorial he was doing. The question is word for word. Jun 16 '20 at 20:07
• Subtracting ${f(1)}$ from both sides gives ${f(1)=0}$, as Savio said. A lot of "paradoxes" rely on the fact that the only solution to this equation is ${0}$ to try and prove ${1=2}$ (notice dividing through by ${f(1)}$ yields ${1=2}$ - but this is not allowed, since we have not checked ${f(1)}$ is $0$, which it is - so maths is not broken) Jun 16 '20 at 20:20

Cool, so as we can see $${f''(x) = \frac{d}{dx}\left(2x^4f(x^2)\right)=8x^3f(x^2)+4x^5f'(x^2)}$$, but $${f'(x)=2x^4f(x^2)}$$ and hence $${f'(x^2)=2x^8f(x^4)}$$, and so overall
$${f''(x)=8x^3f(x^2)+4x^5\left(2x^8f(x^4)\right)=8x^3f(x^2)+8x^{13}f(x^4)}$$
But yeah, I'm not sure what relevance $${f(1)=2}$$ has in this particular case
• It seems to be right, though I wonder what relevance $f(1)$ = 2 had? Jun 16 '20 at 20:41