TITLE: Derivative of $f(x)=e^7+\ln(4)$.
QUESTION [5 upvotes]: I need to differentiate $$f(x)=e^7+\ln(4).$$
I know that $\dfrac d{dx}e^x = e^x$ and $\dfrac d{dx}\ln(x) = \dfrac {1}{x}$. I recently learned this, but I get stuck when it comes to solving this problem using real numbers. Can someone guide me with an example?

REPLY [5 votes]: Consider this alternative. 

Find $g'(x)$ given
  $$g(x) = 2^7 +\log_4(4)$$

Perhaps here you more readily identify that $2^7 = 128$ and $\log_4(4)=1$. That is, both terms are constants: they're numbers that do not depend on the input variable $x$. Now since 
$$g(x) = 129,$$
the derivative of this constant function is zero,
$$g'(x)=0.$$
Your problem is the same, except it involves the irrational number $e=2.71828\dots$ and its logarithm. Perhaps the difficulty is recognizing that $e$, as a symbol, represents a real number. It's like $\pi$ or $\sqrt{2}$, in that it is easiest to represent the number with a symbol rather than work with an interminable decimal or some alternative definition. Similarly, it seems you may have faced some difficulty distinguishing the constant number $\ln(4)$ from the function $\ln(x)$.
Here's one more example:

$$h(x) = \sin\left(\frac{\pi}{3}\right) + \log(57) + \sqrt{3} + 9^{1/7} + 2^e + \pi^\pi$$

Do you see any variables present on the right hand side? There are none. This function is constant. Actually, it's roughly equal to fifty. And since it is constant, $h'(x)=0$.