A curve has equation x = (y+5)ln(2y-7); (i) Find dx/dy in terms of y; (ii) Find the gradient of the curve where it crosses the y-axis.

(i) To find the derivative we will use the product rule. Let u = y+5 and v=ln(2y-7). Then dx/dy = du/dyv + udv/dy = ln(2y-7) + (y+5)*2/2y-7 (used the chain rule in 2nd term - can explain this on white board)(ii) Curve crosses y-axis when x=0. This happens when y=-5 or y=4. y=-5 is not valid since we will get ln(-17) which isn't possible. Plugging y=4 we get that dx/dy = 18, and hence the gradient which is defined as dy/dx = 1/18.