Why is the derivate of $\ln(4) = 0$, but $\lim _{h \rightarrow 0} \frac{\ln(4+h) - \ln(4)}{h}= \frac{1}{4}$

Question

I thought that was the definition of the derivative, but it seems that you can use L'Hospital's rule when the derivative is written with the formal definition way. How is this possible and what am I missing?

Answer 1

When talking about derivatives, you need to think about what function is being differentiated.

Like all constant functions, the derivative of the constant function $f(x) = \ln(4)$ is the function $f'(x) = 0$. So in this case, $f'(4) = 0$.

The derivative of the natural logarithm function $g(x) = \ln(x)$ is $g'(x) = 1/x$. In this case, $g'(4) = 1/4$.

Answer 2

Apparently if you evaluate any function at a given point and take derivative of the result you get zero because you are differentiating a constant.

On the other hand if you take derivative and evaluate the derivative at a given point $a$ you get what is $f'(a)$

Thus $(f(a))'=0$ while $f'(a)$ may or may not be zero.

Answer 3

The limit expression comes directly from the definition of the derivative:

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

To solve this question you must identify two things:

1) What is the particular function $f(x)$ that has been plugged into the expression above?

2) At what point $x = a$ is the derivative being evaluated at?

In this case the function is $f(x) = \ln x$ and it's being evaluated at $x = 4$. That is, $f'(4)$ where $f(x) = \ln x$. From the derivative rule for logarithms, we have

$$f'(x) = \frac{d}{dx} \ln x = \frac{1}{x}.$$

Hence, the limit is $f'(4) = \frac{1}{4}$.

On the other hand, $\ln(4)$ is a constant function, i.e. it's just a number. In this case, it's approximately $1.386$. The derivative of any constant function is zero.

Best of luck.