Why is this the first step of differentiating [a^u(x)]

Question

Following this enter link description here, where we're performing the derivative $$\frac{d}{dx}5^{x\cos(x)},$$

The first step of the differentiation is:

$[a^u(x)]'$ = $ \ln(a) \,a^u(x)\, u'(x).$

I’m quite confused by the intuition. Where did the $\ln$ come from? Why do we need to multiply by $a^u(x)?$

Would appreciate a step by step breakdown of why this is the first step. Is this just another differentiation rule to remember (like chain rule, product rule) that I should instinctively know when performing differentiation, or is there some manipulation that lead to the equation above? This is a new topic for me, apologies.

Answer 1

We have that

$$a^{u(x)}=e^{u(x)\cdot \log a}$$

and then

$$(a^{u(x)})'=(e^{u(x)\cdot \log a})'=e^{u(x)\cdot \log a}\cdot (u(x) \cdot \log a)'=a^{u(x)}\cdot u'(x) \cdot \log a$$

Answer 2

That rule is obtained using logarithmic differentiation. Taking logs essentially converts an exponentiation into a multiplication, allowing you to use the already-familiar product rule. You can use it to derive the generalized power rule for derivatives: $$\left(f^g\right)' = gf^{g-1}\cdot f' + f^g\log f\cdot g'$$ where $f$ and $g$ are functions of $x$ and $'$ denotes differentiation with respect to $x$.

The beauty of this rule is that if $g$ is constant then $g'$ vanishes and you recover the power rule, and if $f$ is constant then $f'$ vanishes and you recover the exponential rule. That makes it really easy to memorize.

To derive it using logarithmic differentiation, write

$$y=f^g$$ We're looking for $y'$. Taking logs: $$\log y = g\log f\tag{power property for logs}$$ $$\tfrac{y'}{y}=g\cdot\tfrac{f'}{f}+\log f\cdot g'\tag{prod rule, deriv of log}$$ $$y'=gy\cdot\tfrac{f'}{f}+y\log f\cdot g'\tag{mult by $y$}$$ $$y'=g(f^g)(f)^{-1}\cdot f' +(f^g)\log f\cdot g'\tag{defn of $y$}$$ $$y'=\boxed{gf^{g-1}\cdot f' + f^g\log f\cdot g'}\tag{combine exponents}$$