# Thread: Chain rule - 2 variables

1. ## Chain rule - 2 variables

Hello MHF,
Would appreciate some help with the following question,
Find $\displaystyle \frac{dz}{dt}$ if $\displaystyle z=txy^2 ,\ x=t+\ln(y+t^2)\ \mbox{and}\ y=e^2$. You may express your answer as a function of $\displaystyle t,x,y.$

2. Originally Posted by Robb
Hello MHF,
Would appreciate some help with the following question,
Find $\displaystyle \frac{dz}{dt}$ if $\displaystyle z=txy^2 ,\ x=t+\ln(y+t^2)\ \mbox{and}\ y=e^2$. You may express your answer as a function of $\displaystyle t,x,y.$
Using some substitutions

$\displaystyle z=txy^2$ becomes $\displaystyle z=t(t+\ln(e^2+t^2))(e^2)^2$

Now you only have to find $\displaystyle \frac{dz}{dt}$ with respect to 1 variable. I would expand from here and differentiate each term separately.

3. In general: $\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$ if $\displaystyle z(x,y)=z(x(t),y(t))$

4. Actually, here, your original function z is a function of three variables, x, y, and t.

The chain rule for this would be $\displaystyle \frac{dz}{dt}= \frac{\partial z}{\partial t}+ \frac{\partial z}{\partial x}\frac{dx}{dt}+ \frac{\partial z}{\partial y}\frac{dy}{dt}$

The $\displaystyle \frac{\partial z}{\partial t}$, of course, would just be $\displaystyle xy^2$.

5. Good catch, I didn't notice that the function was made of three variables.