Why is this partial derivative different if I apply an exponent law prior to taking it?

Question

So I'm solving the following Lagrangian: $L = h + k + \lambda(q^* - 3k^{A^{1/4}}h^{A^{3/4}})$, the first step of which is taking the partial with respect to $k$. Now, I can rewrite the parenthesesed expression as $(q^* - 3k^{A/4} h^{3*A/4})$ using an exponent law. If I take take the partial wrt k of the first expression I get $1 = 3\lambda A^{1/4}k^{A^{1/4} - 1} h^{A^{3/4}} $. If I do it to the second expression though, I get $1 = \frac34 \lambda Ak^{A/4 - 1} h^{3A/4 - 1}$. When I continue on, taking the partial wrt to h, and then divide the two to cancel out lambda, I either end up with $1 = A^{1/2}\frac{k}{h}$ or $1 = 3\frac{k}{h}$, which are both different. What am I doing wrong here? It seems that taking the partial after applying the exponent law and before gives different results?

Answer 1

Hint: The exponent law is $\left(a^b\right)^c=a^{b\cdot c}$. But you don´t have brackets here. Therefore the first partial derivative w.r.t. $k$ is

$1-\lambda \cdot 3\cdot A^{1/4}\cdot k^{A^{1/4}-1}\cdot h^{A^{h/4}}=0\Rightarrow 1=\lambda \cdot 3\cdot A^{1/4}\cdot k^{A^{1/4}-1}\cdot h^{A^{3/4}}$.

The first partial derivative w.r.t. $h$ is

$1-3\cdot \lambda \cdot A^{3/4}\cdot k^{A^{1/4}}\cdot h^{A^{3/4}-1}=0\Rightarrow 1= \lambda \cdot 3\cdot A^{3/4}\cdot k^{A^{1/4}}\cdot h^{A^{3/4}-1}$

Dividing first equation by the second equation.

$$1=A^{-2/4}\cdot \frac{h}{k}$$