Sorry if this should be in the Calculus section, it still shows up as having malware.

The question:

A 5m ladder is leaning against a vertical wall. Suppose that the bottom of the ladder is being pulled away from the wall at a rate of 1m/s. How fast is the area of the triangle underneath the ladder changing at the instant that the top of the ladder is 4m from the floor?

My attempt:

A = (1/2)xy

dA/dt = dA/dx + dx/dt

dA/dt = (1/2)(dx/dt)y + (1/2)(dy/dt)x

Sub in y = 4, x = 3 (using trig),

dA/dt = 3.5(dy/dt)

I'm not sure if I'm doing this correctly, it doesn't seem right that I'd have to solve for dy/dx. The answer is supposed to be 7/8.

Any assistance would be great!