$\lim_{x \to 5} \frac{f(x^2)-f(25)}{x-5}$
Assuming that $f$ is differentiable for all $x$, simplify.
(It does not say what $f(x)$ is at all)
My teacher has not taught us any of this, and I am unclear about how to proceed.
$\lim_{x \to 5} \frac{f(x^2)-f(25)}{x-5}$
Assuming that $f$ is differentiable for all $x$, simplify.
(It does not say what $f(x)$ is at all)
My teacher has not taught us any of this, and I am unclear about how to proceed.
$ \frac{f(x^2)-f(25)}{x-5} = \frac{f(x^2)-f(25)}{x^2-25} \cdot (x+5)$
Since, $f$ is differentiable, if $x\to 5$ then $x^2\to 25$, so taking the lim will give you $f'(25)\cdot 10$.
Apply Lagrange theorem to $f(x^2)$
$f$ is differentiable, so $g(x) = f(x^2)$ is also differentiable. Let's find the derivative of $g$ at $x = 5$ using the definition.
$ g'(5) = \lim_{x \to 5} \frac{g(x) - g(5)}{x - 5} = \lim_{x \to 5} \frac{f(x^2) - f(25)}{x - 5} $
Now write $g'(5)$ in terms of $f$ to get the desired result.