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