1
$\begingroup$

I have the following problem: $f(x, y) = x e^{x^2 + y^2}$, $c(t) = (t, −t)$, and I need to find $(f \circ c)'(t)$.

What I did was take the derivative of $f(x,y)$ with respect to $x$ and $y$, and I got the following gradient: $Df(x,y)=[2(x^2+1)(e^{x^2 + y^2}), 2xye^{x^2 + y^2}]$

I then took $c'(t)$ and got $[1, -1]$ (but as a vertical matrix, I don't know how to type that on here).

I substituted $x=1$, $y=-1$ and got $e^2$, but my grading program says it isn't correct. Can someone help me figure out where I might have gone astray?

$\endgroup$
1
  • $\begingroup$ A negative times a negative is a positive for 2xy for x=1, y=-1 is -2 and you multiply the second entry by negative one so you get a positive and should get about five e squared I believe. $\endgroup$ Commented May 6, 2017 at 3:50

1 Answer 1

Reset to default
1
$\begingroup$

Instead of trying to do multivariate chain rule, just compose the functions together first to get a standard scalar function, then differentiate afterwards: $$ (f \circ c)(t) = f(t, -t) = te^{t^2 + (-t)^2} = te^{2t^2} \implies (f \circ c)'(t) = 4t^2 e^{2 t^2} + e^{2 t^2} $$

Otherwise, if you want to use multivariate chain rule, then let $g(t) = (f \circ c)(t) = f(x(t), y(t))$ and observe that: \begin{align*} \frac{dg}{dt} &= \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} \\ &= (2 x^2e^{x^2 + y^2} + e^{x^2 + y^2})(1) + (2xye^{x^2 + y^2})(-1) \\ &= (2 t^2e^{t^2 + (-t)^2} + e^{t^2 + (-t)^2}) - (2t(-t)e^{t^2 + (-t)^2}) \\ &= (2t^2e^{2t^2} + e^{2t^2}) + (2t^2e^{2t^2}) \\ &= 4t^2e^{2t^2} + e^{2t^2} \\ \end{align*}

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.