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?

  • $\begingroup$ 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 $).$ $\endgroup$ – Lucian Sep 12 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*}

  • $\begingroup$ 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 $\endgroup$ – RandomAspirant Sep 12 at 17:23
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    $\begingroup$ Are you arguing this is a good mnemonic for what to do, or a reliably true bit of algebra that works always? $\endgroup$ – Yakk Sep 12 at 19:39
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    $\begingroup$ 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 $\endgroup$ – Theo Bendit Sep 12 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 $x\in I$

  • $\begingroup$ can you express this in terms of $a(x)^{b(x)} = e^{b(x)ln(a(x))}$? $\endgroup$ – Yakk Sep 12 at 19:41
  • $\begingroup$ Sure, I'm saying fold that into your answer. Note that $e^{b(x) ln a(x)}$ can be simplified as $a(x)^{b(x)}$ $\endgroup$ – Yakk Sep 12 at 19:46
  • $\begingroup$ What is $\mathbb R^{+*}$ ? I've seen $\mathbb R^+$ and $\mathbb R^*$ before, but not this. $\endgroup$ – mr_e_man Sep 12 at 20:38
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    $\begingroup$ 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 $\pi^{\sqrt{2}}$, $\gamma^{e}$... $\endgroup$ – Tuvasbien Sep 14 at 15:20
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    $\begingroup$ @EricDuminil math.stackexchange.com/q/1657619/79767 $\endgroup$ – djechlin Sep 15 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.


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