Derivative of $f(x)=e^7+\ln(4)$. 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?
 A: 
Well, the law says that if $f(x)=c$ where $c$ is any constant then $$f'(x)=0.$$ 

In your question, $$f(x)=e^7+\ln 4$$ where the right hand side is clearly a constant. Thus, $$f'(x)=0.$$
A: 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, $h'(x)=0$.
A: I get the feeling that there is a slight confusion why $f(x)$ is constant. What does it mean that $f$ is constant? It means that if $x$ changes, then $f(x)$ remains the same. As you can see in the definition of $f(x)$, nothing depends on $x$, it is always equal to $e^7 + \ln 4$. Hence we say that $f$ is constant. This means that the derivative with respect to $x$ is $0$, i.e.
$$f'(x) = 0.$$
A: By definition,
\begin{align}
f'(x) &=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
\end{align}
If $f(x)$ is constant or $f(x)=c$ then
\begin{align}
f'(x) &=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\
 &=\lim_{h\to 0}\frac{c-c}{h}\\
&=\lim_{h\to 0}\frac{0}{h}\\
&=0
\end{align}
A: Perhaps you would be able to better visualize the problem at hand using a plot of $f(x)=e^7+ln(4)$.
 
As you can see, the curve is constant and does not change when you change the value of $x$ (or "travel" along the horizontal axis).
Now let's come to differentiation. What a derivative of a function shows is how $f(x)$ changes with $x$. More specifically, it gives you the slope at a specific point, meaning, if you change $x$ by a very small quantity ∆$x$, how much that would change the value of $f(x).$  
In this case, you can straightaway see that when you change $x$, the value of the function, $y$, will not change. In other terms, $y$ will change by $0$, hence the answer you're looking for is $0$.
A: The natural log of $4$, which is denoted $\ln{4}$, is a constant. Therefore, its derivative is $0$. 
Likewise, $\exp{7}= e^7$ is a constant.
