How to do the derivative when an exponent has an exponent

Question

I am trying to solve an equation that is in the form of $y(x) = (c + x^2)^{x^2}$. Note $c =$ constant

My initial thoughts are I need to look into using ln and e to solve this. However what I am really unsure how to deal with the exponent on the right of the equation because it has an exponent as well.

i.e the part highlighted in bold confuses me.

Answer 1

Note $\ln y = (x^2) \ln {(c + x^2)}$. Then $$\frac{dy}{dx} \frac{1}{y} = \frac{d}{dx} ((x^2) \ln {(c + x^2)}) \implies \frac{dy}{dx} = y \frac{d}{dx} ((x^2) \ln {(c + x^2)}) = (c + x^2)^{x^2} \frac{d}{dx} ((x^2) \ln {(c + x^2)})$$

Answer 2

One way to deal with it is to take the exponent out by taking a logarithm:

$$\ln(y) = x^2 \ln \left ( c + x^2 \right ).$$

Now when you differentiate, you get $\frac{y'}{y}$ on the left side, and you have something which is not too hard to differentiate on the right side.

This is called logarithmic differentiation. It's a common trick for differentiating complicated products/quotients, and is essentially the only way to differentiate functions like $x^x$ or your function.

Answer 3

You can also use the more generalized rule $$\left(f^g\right)'=gf^{g-1}\cdot f' + f^g\log f\cdot g'$$ where $f$ and $g$ are functions of $x$ and the prime denotes differentiation with respect to $x$ as usual.

This simultaneously generalizes the familiar power rule and exponential rule. Take $g$ or $f$ to be constant to recover the familiar rules.

You can derive this rule via logarithmic differentiation.