Differentiation of 5^x I differentiated $5^x$ and got $x5^{x-1}.$ 
But the answer is $5^x\log_e 5.$ 
Why so?
 A: $5^x=e^{x\log(5)}$ 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 $(g\circ  f)'(x)=g'(f(x)).f(x)=e^{ax}.a$
A: 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 $5 \times 5^{x-1} \equiv 5^x$
A: 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.
