# Thread: proving differentiablity

1. ## proving differentiablity

Show that if $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable at $x=1$,

$\lim_{x\to 1}{\frac{f(x^2)-f(1)}{x^2-1}}=f'(1)$

How do we go about proving it? Thanks in advance.

2. ## Re: proving differentiablity

Hint By definition, $f'(1)=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1}$ .

3. ## Re: proving differentiablity

Originally Posted by FernandoRevilla
Hint By definition, $f'(1)=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1}$ .
$t\rightarrow{1}\Rightarrow{t^2}\rightarrow{1}$, but $t^2\rightarrow{1}\nRightarrow{t}\rightarrow{1}$, so is it right to say
$f'(1)=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1}=\lim_{t^2\to 1}\frac{f(t^2)-f(1)}{t^2-1}=\lim_{t\to 1}\frac{f(t^2)-f(1)}{t^2-1}$ ?

4. ## Re: proving differentiablity

Originally Posted by alphabeta89
$t\rightarrow{1}\Rightarrow{t^2}\rightarrow{1}$, but $t^2\rightarrow{1}\nRightarrow{t}\rightarrow{1}$, so is it right to say
$f'(1)=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1}=\lim_{t^2\to 1}\frac{f(t^2)-f(1)}{t^2-1}=\lim_{t\to 1}\frac{f(t^2)-f(1)}{t^2-1}$ ?
Use the substitution $t=\sqrt{x}$ so, $\lim_{x\to 1}\frac{f(x^2)-f(1)}{x^2-1}=\ldots$

5. ## Re: proving differentiablity

Originally Posted by FernandoRevilla
Use the substitution $t=\sqrt{x}$ so, $\lim_{x\to 1}\frac{f(x^2)-f(1)}{x^2-1}=\ldots$
So is it correct to say that
$\lim_{x\to 1}\frac{f(x^2)-f(1)}{x^2-1}=\lim_{\sqrt{t}\to 1}\frac{f(t)-f(1)}{t-1}=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1}=f'(1)$?