# Thread: Simple chain rule problem

1. ## Simple chain rule problem

The question:

Let $z = \sqrt{x^2 + y^2}, x = t^2, y = t^3$. Use a chain rule to ﬁnd dz/dt.

My attempt:
$\frac{dz}{dt} = \frac{\partial{z}}{\partial{x}}\frac{dx}{dt} + \frac{\partial{z}}{\partial{y}}\frac{dy}{dt}$

$\frac{\partial{z}}{\partial{x}} = \frac{x}{\sqrt{x^2 + y^2}}$
$\frac{\partial{z}}{\partial{y}} = \frac{y}{\sqrt{x^2 + y^2}}$
$\frac{dx}{dt} = 2t$
$\frac{dy}{dt} = 3t^2$

So we get:
$\frac{t^2}{\sqrt{t^4 + t^6}}(2t) + \frac{t^3}{\sqrt{t^4 + t^6}}(3t^2)$
= $\frac{t^2(2t + 3t^3)}{t^2\sqrt{1 + t^2}}$
= $\frac{2t + 3t^3}{\sqrt{1 + t^2}}$

$\frac{2t^2 + 3t^4}{\sqrt{1 + t^2}}$

What have I done incorrectly? Thanks.

2. You're overcomplicating this...

$\displaystyle z = \sqrt{t^4 + t^6}$.

Let $\displaystyle u = t^4 + t^6$ so that $\displaystyle z = u^{\frac{1}{2}}$.

Go from here.

3. I'd say you are right. An easy way to check is to substitute x(t) and y(t) into the original equation at the start and find dz/dt directly.

4. Ahh yes, that is a lot easier. I guess I got caught up with all this partial derivative stuff at uni.