How could I solve this chain rule problem?

Question

To find: $\frac{d}{dt}[f(\textbf{c}(t))]\rvert_{t=0}$

Where $\textbf{c}(t)$ is such that $\frac{d}{dt}\textbf{c}(t) = \textbf{F}(c(t))$, with $\textbf{F} = \nabla f$, $\textbf{c}(0) = (-\frac{\pi}{2}, \frac{\pi}{2})$ and $f(x,y) = \sin(x + y)xy$

I think I would need to proceed using the chain rule but I am currently not sure. How would I go about solving this problem?

Thanks in advance

Answer 1

Let $g=f \circ c$, i.e. let $$g(t)=f(c(t))=f(c_1(t),c_2(t)).$$ Then by the chain rule

$$g'(t)=f_x(c(t))c_1'(t)+f_y(c(t))c_2'(t)=\nabla f(c(t)) \cdot c'(t)=\nabla f(c(t)) \cdot \nabla f(c(t)),$$

where the second equality follows from the assumption about $c'(t)$.

Since

$$f_x(x,y)=y[\sin(x+y)+x\cos(x+y)]$$

and

$$f_y(x,y)=x[\sin(x+y)+y\cos(x+y)]$$

we have

$$f_x(c(0))=f_x\left(-\frac{\pi}{2},\frac{\pi}{2}\right)=-\frac{\pi^2}{4}$$

$$f_y(c(0))=f_y\left(-\frac{\pi}{2},\frac{\pi}{2}\right)=-\frac{\pi^2}{4}$$

or

$$\nabla f(c(0))=\left(-\frac{\pi^2}{4},-\frac{\pi^2}{4}\right).$$

Thus

$$g'(0)=\nabla f(c(0)) \cdot \nabla f(c(0)) = \left(-\frac{\pi^2}{4},-\frac{\pi^2}{4}\right)\cdot \left(-\frac{\pi^2}{4},-\frac{\pi^2}{4}\right)=\frac{\pi^4}{8}.$$