# Proof derivative equals zero?

I know this must be wrong, but I am confused as to where the mathematical fallacy lies.

Here is the 'proof':

$$f '(x) = \lim_{ h\to0}\frac{f(x+h)-f(x)}{h}$$

L'Hôpital's Rule (The previous limit was $\frac{0}{0}$):

$$f '(x) = \lim_{ h\to 0}\frac{f '(x+h)-f '(x)} {1}$$

Plugging in $h$:

$$f '(x) = f '(x+0)-f '(x)$$

Simplifying:

$$f '(x) = 0$$

I'm assuming my application of L'Hôpital's rule is fallacious, but it evaluates to an indeterminate form so isn't L'Hôpital's rule still valid?

Taking the derivative with respect to $h$ gives:
$$f'(x) = \lim_{h \rightarrow 0} \frac{f'(x + h)}{1}$$
Since $f(x)$ is constant with respect to $h$.
• Ah, and similarly, taking the derivative with respect to $x$ gives $\frac{f'(x+h) - f'(x)}{0}$, which would be invalid. At the time that I posted the question I wasn't familiar enough to Leibniz's notation to realize there's an implicit variable hidden in the apostrophe, but thanks for the clarification! – Nicholas Pipitone Oct 12 '18 at 17:53
You differentiated the numerator with respect to $x$ but the denominator with respect to $h$.