# What is the derivative of a function of the form $u(x)^{v(x)}$?

So I have a given lets say $$(x+1)^{2x}$$ in addition to $$\frac{\mathrm dy}{\mathrm dx}a^u=a^u\log(a)u'$$. I still have to multiply this by the derivative of the inside function $$x+1$$ correct?

• To evaluate the derivative of an expression of the form $\Big[u(x)\Big]^{v(x)},~$ we must combine the two relevant formulas for the derivatives of $a^{u(x)}$ and $\Big[u(x)\Big]^n.~($ Two related questions $).$ Commented Sep 12, 2019 at 10:02

This is what logarithmic differentiation is for. You start with writing the function as an equation $$y = (x + 1)^{2x},$$ then take the natural log of both sides: $$\ln y = \ln\left[(x + 1)^{2x}\right] = 2x \ln(x+1).$$ We then implicitly differentiate both sides with respect to $$x$$. By chain rule (remember, $$y$$ is a function of $$x$$), the left side comes to $$\frac{1}{y} \cdot y'.$$ The right side can be differentiated as normal: $$\frac{2x}{x + 1} + 2\ln(x + 1).$$ So, \begin{align*} &\frac{1}{y} \cdot y' = \frac{2x}{x + 1} + 2\ln(x + 1) \\ \implies \, &y' = y\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right) \\ \implies \, &y' = (x + 1)^{2x}\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right). \end{align*}

• It should be noted that although logarithm is only defined for positive values, this method is applicable on general for any values inside the logarithm Commented Sep 12, 2019 at 17:23
• Are you arguing this is a good mnemonic for what to do, or a reliably true bit of algebra that works always?
– Yakk
Commented Sep 12, 2019 at 19:39
• To address the comments, when working within $\Bbb{R}$, the conventional domain of $a^b$ is $a > 0$ (and $b \in \Bbb{R}$). This means that $a^b > 0$ always. While it may be observed that real values can be ascribed at certain other isolated points (e.g. when $a < 0$ but $b \in \Bbb{Z}$), many rational and all irrational values of $b$ pose a problem when $a < 0$. Therefore, talking about derivatives when $a < 0$ is tricky when the function is not even properly defined densely around such values. Thus, there is no particular problem with assuming the function is positive and taking the logarithm Commented Sep 12, 2019 at 23:20

Since $$(x+1)^{2x}=e^{2x\ln(x+1)}=e^{u(x)}$$, its derivate is $$u'(x)e^{u(x)}$$. Notice that if $$a:I\longrightarrow\mathbb{R}^{+*}$$ and $$b:I\longrightarrow\mathbb{R}$$, then we define $$a(x)^{b(x)}$$ as $$e^{b(x)\ln a(x)}$$ for all $$x\in I$$

• can you express this in terms of $a(x)^{b(x)} = e^{b(x)ln(a(x))}$?
– Yakk
Commented Sep 12, 2019 at 19:41
• Sure, I'm saying fold that into your answer. Note that $e^{b(x) ln a(x)}$ can be simplified as $a(x)^{b(x)}$
– Yakk
Commented Sep 12, 2019 at 19:46
• What is $\mathbb R^{+*}$ ? I've seen $\mathbb R^+$ and $\mathbb R^*$ before, but not this. Commented Sep 12, 2019 at 20:38
• Indeed, the definition $3^2=e^{3\ln 2}$ is compatible with the idea that $3^2=3\times 3$ but it also allows to compute much more powers such as $\pi^{\sqrt{2}}$, $\gamma^{e}$... Commented Sep 14, 2019 at 15:20
• @EricDuminil math.stackexchange.com/q/1657619/79767 Commented Sep 15, 2019 at 3:47

That whole $$a^u$$ thing works when $$a$$ is a constant, not another expression in terms of $$x$$.

To take the derivative of this, you would have to convert it to $$\displaystyle e^{2x\ln(x+1)}$$ and THEN use the Chain Rule.

This would be $$\displaystyle e^{2x\ln(x+1)}\left(2\ln(x+1)+\frac{2x}{x+1}\right)=(x+1)^{2x}\left(2\ln(x+1)+\frac{2x}{x+1}\right)$$

A nice way to do this is to use a somewhat stronger fact than the other answers: as a function of two variables, the expression $$a^b$$ is differentiable. Simply put, what this means is that, if $$f(x)$$ and $$g(x)$$ are functions, and $$h(x)=f(x)^{g(x)}$$, then $$h'(x)$$ is the sum of how fast this expression changes when we treat $$f(x)$$ as a constant plus how fast it changes when we treat $$g(x)$$ as a constant. You already know how to differentiate polynomials and exponentials, so this suffices - just apply both rules and add them!

Since we know that the derivative of $$f^{g(x)}$$ is $$\log(f)\cdot f^{g(x)}\cdot g'(x)$$ and the derivative of $$f(x)^g$$ is $$g\cdot f(x)^{g-1}\cdot f'(x)$$, we get $$h'(x)=\underbrace{\log(f(x))\cdot f(x)^{g(x)}\cdot g'(x)}_{\text{Derivative treating f as constant}}+\underbrace{g(x)\cdot f(x)^{g(x)-1}\cdot f'(x)}_{\text{Derivative treating g as constant}}.$$ You would get the same result by using logarithmic differentiation as other answers suggest*, but I generally find this is a bit easier to remember and more generalizable - for instance, note that differentiating a product $$f(x)\cdot g(x)$$ can be done by this same method.

(*Logarithmic differentiation is a good way to prove this result, since you write $$\log(h(x)) = \log(f(x))\cdot g(x)$$then differentiate both sides to get $$\frac{h'(x)}{h(x)}=\frac{f'(x)\cdot g(x)}{f(x)}+\log(f(x))\cdot g'(x)$$ and moving $$h(x)$$ to the other side and substituting it for its formula gives the formula I claim)

Start with $$f(x) = a(x)^{b(x)}$$.

Now, we can approach this a few ways. The one way is to go into multidimensional derivatives. Define

$$h(x,y) = a(x)^{b(y)}$$

$$h'(x,y) = \left[ {\begin{array}{cc} \frac{ \partial h_x }{ \partial x } & \frac{ \partial h_x }{ \partial y } \\ \frac{ \partial h_y }{ \partial x } & \frac{ \partial h_y }{ \partial y } \\ \end{array} } \right] = \left[ {\begin{array}{cc} a'(x) b(y) a(x)^{b(y)-1} & 0 \\ 0 & ln(a(x)) b'(y) a(x)^{b(y)}\\ \end{array} } \right]$$

Let $$g(x) = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$$. Then $$f(x) = h(g(x))$$. By the chain rule, $$f'(x) = h'(g(x)) g'(x) = h'(x,x) \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$$

or $$f'(x) = a'(x) b(x) a(x)^{b(x)-1} + ln(a(x)) b'(x) a(x)^{b(x)}$$

$$f'(x) = f(x) * ( b(x) \frac{a'(x)}{a(x)} + b'(x) ln(a(x)))$$

We can then do a sanity check using logarithmic derivatives. Also we can look at what happens when $$a(x) = k$$:

$$f'(x) = f(x) * ( b(x) \frac{0}{a(x)} + b'(x) ln(k))$$ $$f'(x) = ( b'(x) ln(k) ) k^{b(x)}$$

or $$b(x)=k$$:

$$f'(x) = f(x) * ( k \frac{a'(x)}{a(x)} + 0 ln(a(x)))$$ $$f'(x) = k a'(x) a(x)^{k-1}$$

which are both right.

Conceptually, this has two terms. You add them to find out what the instant "percent change" in $$f(x)$$ will be.

$$b \frac{a'}{a}$$ is $$b$$ times the instant percent change in $$a$$: If $$a$$ is growing by $$1\%$$, and you raise this to the power $$b$$, you grow by $$b\%$$.

$$b' ln a$$ is the change in $$b$$ wrt $$x$$ scaled by the log-scale of $$a$$. With exponentials, a given linear-scale unit of exponent increase causes a scale by a factor in the base. $$ln(a)$$ is the ratio between the two (the linear effect in the exponent to the exponential effect in the result), and we multiply that by $$f$$ because we are scaling the entire value.