5
$\begingroup$

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?

$\endgroup$
8
  • 23
    $\begingroup$ $f'(x)=0$, as $f$ is constant. $\endgroup$
    – S.C.B.
    Feb 20, 2017 at 1:06
  • 12
    $\begingroup$ 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
  • 1
    $\begingroup$ @Shufflepants $\dfrac1x$ or if you prefer $\dfrac4{4x}$ $\endgroup$
    – Henry
    Feb 20, 2017 at 23:42

6 Answers 6

29
$\begingroup$

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

$\endgroup$
2
  • 4
    $\begingroup$ 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
  • 7
    $\begingroup$ @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
16
$\begingroup$

Perhaps you would be able to better visualize the problem at hand using a plot of $f(x)=e^7+ln(4)$.

$f(x)=exp(7)+ln(4)$

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 $0$.

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
  • 10
    $\begingroup$ @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
  • 8
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
9
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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, $h'(x)=0$.

$\endgroup$
1
  • $\begingroup$ Students have been observed who have arrived at this result using the chain rule etc. $\endgroup$
    – Carsten S
    Feb 21, 2017 at 2:04
2
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.