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?

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    $\begingroup$ $f'(x)=0$, as $f$ is constant. $\endgroup$
    – S.C.B.
    Feb 20, 2017 at 1:06
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    $\begingroup$ There is no $x$ in your definition of $f(x)$ so the function does not depend on $x$ i.e. it's a constant. $\endgroup$
    – dxiv
    Feb 20, 2017 at 1:12
  • $\begingroup$ I'll try to restate what others have said: $e^7+\ln4$ is just a "number". Like $\pi$, or $8\times7 + 9$. They're all just numbers, constants. The derivative of a constant is $0$ $\endgroup$ Feb 20, 2017 at 17:17
  • $\begingroup$ If you had instead $f(x) = e^x + ln(x)$ then the derivative would have been $f'(x) = e^x + 1/x$ $\endgroup$ Feb 20, 2017 at 19:58
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    $\begingroup$ @Shufflepants $\dfrac1x$ or if you prefer $\dfrac4{4x}$ $\endgroup$
    – Henry
    Feb 20, 2017 at 23:42

6 Answers 6


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. $$f'(x) = 0.$$

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    $\begingroup$ Whew, 21 upvotes for defining the word constant XD got to love how rep works on this site. $\endgroup$ Feb 20, 2017 at 17:55
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    $\begingroup$ @BrevanEllefsen Well, technically no rep because I submitted as community wiki, haha :P. I remember myself not understanding these concepts when I was younger, so I try to put myself into the shoes of my younger self. $\endgroup$
    – Eff
    Feb 20, 2017 at 18:24

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 $0$.


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.

  • $\begingroup$ I knew about the constant rule but didn't know that it could be applied to this. So basically if I have a constant in the front it would always be zero ? $\endgroup$
    – Jeff
    Feb 20, 2017 at 1:10
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    $\begingroup$ @Jeff It's not constant "in the front". It's simply constant. The variable $x$ does not appear in the function. $\endgroup$
    – NoName
    Feb 20, 2017 at 1:28
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    $\begingroup$ As @NoName has said, $e^7$ is just "a number"... you can plug it into a calculator for a decimal approximation (~$1096.6...$). What would the derivative of $1096.6...$ be? $\endgroup$ Feb 20, 2017 at 1:33
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    $\begingroup$ @Jeff Rather than looking for "a constant in the front," the question to ask yourself regarding an expression like $e^7+\ln(4)$ is: where is the variable? $\endgroup$
    – David K
    Feb 20, 2017 at 19:08

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, $$f'(x)=0.$$


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, $h'(x)=0$.

  • $\begingroup$ Students have been observed who have arrived at this result using the chain rule etc. $\endgroup$
    – Carsten S
    Feb 21, 2017 at 2:04

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}


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