Christoffel of 2nd kind equals partial derivitive of ln g_aa

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if metric tensor $g_{ij}$ is a diagonal matrix, show that for all fixed indices (no summation) in the range: 1, 2, ..., n, that:

$\huge \Gamma^\alpha_{\alpha\beta} = \frac{\partial}{\partial x^\beta} \bigg( \frac{1}{2} \ln |g_{\alpha\alpha}|\bigg)$

tensor text book says its:

Step 1: $\huge \Gamma^\alpha_{\alpha\beta} = g^{\alpha j} \Gamma_{\alpha \beta j}$

Step 2: $\huge \Gamma^\alpha_{\alpha\beta} = g^{\alpha \alpha} \Gamma_{\alpha \beta \alpha} $

Step 3: $\huge \Gamma^\alpha_{\alpha\beta} = \frac{1}{g_{\alpha\alpha}}(\frac{1}{2} \frac{\partial g_{\alpha\alpha}}{\partial x^{\beta}})$

Step 4: $\huge \Gamma^\alpha_{\alpha\beta} = \frac{\partial}{\partial x^\beta}(\frac{1}{2} \ln |g_{\alpha\alpha}|)$

I'm ok with step 1-3... But, how did they get from step 3 to step 4? I'm not understanding where Natural Log came from...

(Exercise in quetion is from "Schaum's Outlines: Tensor Calculus", Kay,2011, page 75, problem 6.4)

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If you differentiate $ \ln |g_{\alpha\alpha}|$ with respect to $x_\beta$ you get $ \frac{1}{g_{\alpha\alpha}}(\frac{\partial g_{\alpha\alpha}}{\partial x^{\beta}})$