# Quick chain rule q

1. May 4, 2012

### wumple

Hi,

If I have the equation

$$y' = ax - by$$

where $$y = y(t) , x= x(t)$$

and $$y' = \frac{dy}{dt}$$

then what is

$$\frac {d}{dy} y' = \frac {d}{dy}(ax - by)$$

?

I think it would come out to

$$\frac {dy'}{dy} = a \frac {dx}{dt}\frac {dt}{dy} - b$$

Is that right? In general, is y' a function of y or would the first term on the left be 0?

Thanks!

2. May 5, 2012

### Staff: Mentor

What you have is good, as far as you went, but you can do more with the expression on the left side of the equation.

Using the chain rule, we have
d/dy(y') = d/dy(dy/dt) = d/dt(dy/dt) * dt/dy = y'' * 1/y'