Consider the equation $z^2-1=x^3y$. Find the value of $\frac{dy}{dt}$ under these conditions $z=5,x=2,y=3, \frac{dx}{dt}=-2$ and $\frac{dz}{dt}=7$.
So I'm not really getting this. I think what I do is take the derivative of the original so I get $2z=3x^2$. Is this correct? What I don't get is where to go from there. Not sure what to plug in. Appreciate any help :)
$z^2−1=x^3y\\ 2z\frac {dz}{dt} = 3x^2y \frac {dx}{dt} + x^3\frac{dy}{dt}$
Now plug your known values for $x,y,z, \frac {dx}{dt},\frac {dz}{dt}$ and solve for $\frac {dy}{dt}$
Because we are trying to find $\frac{dy}{dt}$, we have to differentiate everything by $t$. If we do so to both sides, we get:
$\begin{eqnarray} \frac{d \left( z^2 - 1\right) }{dt} & = & \frac{d \left( x^3 y \right) }{dt} \\ \frac{d \left(z^2 - 1\right) }{dz}\frac{dz}{dt} & = & \frac{d\left(x^3\right)}{dt}y + x^3 \frac{dy}{dt} \\ 2z \frac{dz}{dt} & = & 3x^2 \frac{dx}{dt} y + x^3 \frac{dy}{dt}\end{eqnarray}$
In the second line I've applied the chain rule on the left, and the product rule on the right. In the third, I applied the chain rule to the $x$ component.
Given that, you can just put values in and solve, but you should also confirm that you understand how that all works.
