Question

If z=(x+y)e^y and x=5t and y=1−t^2, find the following derivative using the chain rule. Enter your answer as a function of t.

Answer 1

By the chain rule, if $z=f(x,y)=f(x(t),y(t))$, then

$\dfrac{\mathrm dz}{\mathrm dt} = \dfrac{\partial f}{\partial x}\dfrac{\mathrm dx}{\mathrm dt} + \dfrac{\partial f}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

We have $f(x,y)=(x+y)e^y$, for which

• $\dfrac{\partial f}{\partial x} = e^y$

• $\dfrac{\partial f}{\partial y} = e^y + (x+y)e^y = (x+y+1)e^y$

and $x(t)=5t$ and $y(t)=1-t^2$, so that

• $\dfrac{\mathrm dx}{\mathrm dt} = 5$

• $\dfrac{\mathrm dy}{\mathrm dt} = -2t$

Putting everything together, we get

$\dfrac{\mathrm dz}{\mathrm dt} = 5e^y - 2t(x+y+1)e^y \\\\ \dfrac{\mathrm dz}{\mathrm dt} = 5e^{1-t^2} - 2t(5t-t^2+2)e^{1-t^2} = \boxed{\left(2t^3-10t^2-4t+5\right)e^{1-t^2}}$