0
$\begingroup$

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

  • 0
    Please, try to make the title of your question more informative. E.g., *Why does $a imply $a+c?* is much more useful for other users than *A question about inequality.* From [How can I ask a good question?](http://meta.math.stackexchange.com/a/589/): *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.*2012-11-07

3 Answers 3

0

$$\dfrac{df(x^2)}{dx} = \dfrac{df(x^2)}{d(x^2)} \dfrac{d(x^2)}{dx} = 2x \dfrac{df(y)}{dy}$$ 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 $$f(y) = \dfrac{\log(y)}2 +1$$

  • 0
    How did you get what $\frac{1}{x}$ is ?2012-11-07
  • 0
    I read $f'(x^2)$ as the derivative of $f$ evaluated at $x^2$. That is, $f'(x)=\frac1{\sqrt{x}}$.2012-11-07
  • 0
    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.2012-11-07
3

$f '(x^2)=1/x \Rightarrow$ 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

1

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 $$ f(x)=2\sqrt x - 1 $$