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I differentiated $5^x$ and got $x5^{x-1}.$

But the answer is $5^x\log_e 5.$

Why so?

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    $\begingroup$ This is a common mistake. The rule you want to use only works if the base is the variable and the exponent is a number. $\endgroup$ Commented Jun 13, 2018 at 23:29
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    $\begingroup$ The rule for powers applies when you differentiate with respect to the base, not with respect to the power. $\endgroup$ Commented Jun 13, 2018 at 23:30
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    $\begingroup$ You are misusing the rule. The rule you are trying to use applies to functions of the form $x^n$, where $n$ is a constant. Not functions of this form. $\endgroup$
    – lulu
    Commented Jun 13, 2018 at 23:30
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    $\begingroup$ The rule is that if $f(x)=x^n$ for some $n$, then $f'(x)=nx^{n-1}$. Your function is not of this form. Your function is of the form $f(x)=n^x,$ which is entirely different.Part of the problem is that the notation $f'$ doesn't indicate what you are differentiating against. $\endgroup$ Commented Jun 13, 2018 at 23:37
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    $\begingroup$ $x$ is not a constant and $5$ is not a variable. $x^n$ and $n^x$ are very different function. And a small change in the value of $x$ has different effects. $\frac {(x + h)^5 -x^5}{h}=\frac {x^5 + 5hx^4 + 10h^2x^3 + 10h^3x^2 + 5h^4x + h^5 - x^5}h = 5x^4 + 10hx^3+10h^2x^2+5h^3x + h^4\to 5x^4$ where as $\frac {5^{x+h} - 5^{x}}{h} = \frac {5^x*5^h -5^x}h= \frac {5^x(5^h -1)}{h$ which behaves very differently. $\endgroup$
    – fleablood
    Commented Jun 13, 2018 at 23:41

3 Answers 3

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$5^x=e^{x\log(5)}$ so its differential is $\log(5)e^{x\log(5)}=\log(5)5^x$.

The differential of $e^{ax}=ae^{ax}$, write $f(x)=ax, g(x)=e^x$, $f'(x)=a, g'(x)=e^x$, you have $(g\circ f)'(x)=g'(f(x)).f(x)=e^{ax}.a$

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    $\begingroup$ @LianaA typically if mathematicians write $\log$, they mean $\ln$, so yes :) $\endgroup$ Commented Jun 13, 2018 at 23:39
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    $\begingroup$ $\log_{10}$ is nearly completely useless now that we no longer do multiplication using logarithm tables out of a book or slide rules. :) $\log_2$ is more often useful. $\endgroup$ Commented Jun 13, 2018 at 23:46
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    $\begingroup$ @Robert Lewis Interesting, thank you for point it out. $\endgroup$
    – AnyAD
    Commented Jun 13, 2018 at 23:50
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    $\begingroup$ @RobertLewis My profile picture used to be a photo of a statue of Bowditch in Mount Auburn Cemetery, where he is buried (right across the street from where I live.) flic.kr/p/pzh3jx $\endgroup$ Commented Jun 14, 2018 at 0:56
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    $\begingroup$ “Derivative”, not “Differential” $\endgroup$
    – MPW
    Commented Jun 14, 2018 at 2:03
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For a generalized answer to this question, you can use the following which works in cases of $5^x$ and $x^5$:

$$ \frac{\mathrm{d}}{\mathrm{d}x}\left(f(x)^{g(x)}\right) = f(x)^{g(x)-1}(g(x)f'(x)+f(x)\log(f(x))g'(x) $$

So

$$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}x^5 &= x^{5-1}(5 \times 1 + x \log(x) \times 0) \\ &= x^4(5+0) \\ &= 5x^4 \end{align} $$

And, likewise:

$$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} 5^x &= 5^{x-1}(x \times 0 + 5 \log(5) \times 1) \\ &= 5^{x-1}(0 + 5 \log(5)) \\ &= 5^{x} \log(5) \end{align} $$

Note $5 \times 5^{x-1} \equiv 5^x$

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One needs to respect two things:

first, the formulas

$\dfrac{dx^n}{dx} = nx^{n - 1}, \; n \ge 1 \tag 1$

and

$\dfrac{du^n(x)}{dx} = nu^{n - 1}(x)\dfrac{du}{dx}, \; n \ge 1, \tag 2$

where $u$ is a differentiable function of $x$, only apply when $n$ is a constant and the variable ($x$ or $u(x)$ here) occurs in the base, not in the exponent;

second, when the base is constant and the exponent varies, we are really dealing with a case of $e^{u(x)}$, which by the chain rule satisfies

$\dfrac{de^{u(x)}}{dx} = e^{u(x)} \dfrac{du(x)}{dx}; \tag 3$

in the present instance, we have

$5^x = (e^{\ln 5})^x = e^{(\ln 5)x}, \tag 4$

so

$\dfrac{d(5^x)}{dx} = \dfrac{de^{(\ln 5)x}}{dx} = e^{(\ln 5)x} \dfrac{d(\ln 5)x}{dx} = \ln 5 e^{(\ln 5)x} = (\ln 5)(5^x); \tag 5$

it is also worth noting that the formula (1), for $n \in \Bbb Z_+$, follows from the binomial theorem; for example, with $n = 3$, we have

$(x + h)^3 - x^3 = x^3 + 3x^2 h + 3x h^2 + h^3 - x^3 = 3x^2 h + 3xh^2 + h^3, \tag 6$

whence

$\dfrac{(x + h)^3 - x^3}{h} = 3x^2 + (3xh + h^2) \to 3x^2 \; \text{as} \; h \to 0, \tag 7$

with a similar derivation for any $n \in \Bbb Z_+$; such an approach clearly won't work for functions of the form $a^x$, where we have

$\dfrac{a^{x + h} - a^x}{h} = a^x\dfrac{a^h - 1}{h}, \tag 8$

and thus we need to prove

$\displaystyle \lim_{h \to 0} \dfrac{a^h - 1}{h} = \ln a, \tag 9$

which follows from

$a^h = e^{(\ln a) h} = \displaystyle \sum_0^\infty \dfrac{(\ln a)^n h^n}{n!}. \tag{10}$

I leave the elementary details, which involve some notions of power series' convergence, to my readers.

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