Gradient of $\ln(y^2)=\frac{1}{2}x\ln(x-1)$ at point $x=4$

Question

What is the gradient of the curve:

$\ln(y^2)=\frac{1}{2}x\ln(x-1)$

$x>1, y>0$ when $x = 4$.

How would I differentiate this equation?

Then I presume you just sub 4 in as x to find the gradient.

Answer 1

It is $$\frac{1}{y^2}2yy'=\frac{1}{2}\ln(x-1)+\frac{1}{2}x\times\frac{1}{x-1}$$

Answer 2

You may rearrange the function first,

$$\ln y = \frac14 x\ln(x-1)$$

and differentiate,

$$\frac{y'}{y} = \frac14\left(\ln(x-1)+\frac{x}{x-1}\right)$$

Then, plug in $x=4$ to get $y(4)=\ln3$ and

$$y'(4)=\frac{\ln3}{4}\left(\ln3+\frac43\right)$$