# Hyperbolic Differentiation

1. Oct 27, 2015

### DiamondV

1. The problem statement, all variables and given/known data
Differentiate

2. Relevant equations
Chain Rule: dg/dx = du/dx . dv/du . dg/dv

3. The attempt at a solution

Correct answer provided to us(not mine):

I understand the correct solution that was provided to us, but what I don't understand is why my method isnt correct? Also can you check in particular my dv/du . I suspect theres something wrong there.

2. Oct 27, 2015

### pasmith

The derivative of $3^v = e^{v \log 3}$ is $(\log 3) e^{v \log 3} = (\log 3)3^v$, not $3v^{v-1}$ which appears to be what you have.

If you set $v = u\log x$ then you want to calculate $$\frac{dv}{du} = \log x + \frac ux \frac{dx}{du}$$ and the $\frac{dx}{du}$ cancels with $\frac{du}{dx}$ when you apply the chain rule.

Differentiating functions in exponents is straightforward: By the chain rule, with $a > 0$, $$\frac{d}{dx} a^{f(x)} = (\log a)a^{f(x)} f'(x).$$