# Determine $f$ such that $f'(x^2)=\frac{1}{x}$ for $x>0$ and $f(1)=1$

I am having a problem with the following exercise. I have to determine the function f such that:

$$f'(x^2)=\frac{1}{x} \text{ for } x>0, \quad f(1)=1$$

Thank you in advance

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Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Julian Kuelshammer Nov 7 '12 at 20:19

$$\dfrac{df(x^2)}{dx} = \dfrac{df(x^2)}{d(x^2)} \dfrac{d(x^2)}{dx} = 2x \dfrac{df(y)}{dy}$$ where $y=x^2$. Hence, we get that $$\dfrac1{\sqrt{y}} = \dfrac1{x} = \dfrac{df(x^2)}{dx} = 2\sqrt{y} \dfrac{df(y)}{dy}$$ Hence, we have $$\dfrac{df(y)}{dy} = \dfrac1{2y}$$ This gives us that $f(y) = \dfrac{\log(y)}2 + C$. Since $f(1) = 1$, we get that $C=1$. Hence, we have $$f(y) = \dfrac{\log(y)}2 +1$$

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How did you get what $\frac{1}{x}$ is ? – Belgi Nov 7 '12 at 20:34
I read $f'(x^2)$ as the derivative of $f$ evaluated at $x^2$. That is, $f'(x)=\frac1{\sqrt{x}}$. – robjohn Nov 7 '12 at 20:41
Marvis, note the OP wrote $f'(x^2)$ and not $f(x^2)'=f'(x^2)2x$ You're interpreting this as the latter, but I think it is not the case. – Pedro Tamaroff Nov 7 '12 at 20:51

We have $xf'(x^2)=1$. Let $g(x)=f(x^2)$. Then $g'(x)=2xf'(x^2)$. So the original condition reads $$g'(x)=2.$$ This tells us that $g(x)=2x+c$ for some constant $c$. But then, as we consider positive $x$, $$f(x)=f((\sqrt x)^2)=g(\sqrt x)=2{\sqrt x}+c.$$ As $f(1)=1$, we have $1=2+c$, so $c=-1$. Then $$f(x)=2\sqrt x - 1$$

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$f '(x^2)=1/x \Rightarrow$ if you only plug in $x$ to the function $f '$ this yields the mentioned $f '(x)=x^{-1/2}$.

integrating this gives you $f(x)=2x^{1/2}+C$.

now adjust the const so that $f(x)=1$ where $x=1$. (you might find something like -1...)

Cheers Fab

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