Differentiating a function of x and y with respect to t if x and y are both functions of t.

Question

Suppose I have a function such as: $$f(w) = e^{2w(x,y)}$$

where x and y are bot a function of t.

How would I calculate then the following derivative:

$$\frac{df(w)}{dt}$$

Answer 1

Write $g(t)=e^{2t},\ \gamma(t)=(x(t),y(t))$ and $\omega:\mathbb{R^2}\to \mathbb{R}$. Then, $$f(t)=g(\omega(\gamma(t)))=e^{2\omega(x(t),y(t))}$$

Applying the chain rule once we get $$\tag{1}\frac{\partial}{\partial t}f(t)=J_g|_{\omega(\gamma(t))}\cdot J_{\omega\circ\gamma}|_{t}$$

where $J_g|_{\omega(\gamma(t))}$ is the Jacobian of $g$ evaluated at $\omega(\gamma(t))$ and $J_{\omega\circ\gamma}|_{t}$ is the Jacobian of $\omega\circ \gamma$ evaluated at $t$. The Jacobian of $g$ at $\omega(\gamma(t))$ is just $e^{2\omega(\gamma(t))}$. Then applying chain rule again to evaluate $J_{\omega\circ\gamma}|_{t}$ we have $$J_{\omega\circ\gamma}|_{t}=J_{\omega}|_{\gamma(t)}\cdot J_\gamma|_t$$

Now, since $\omega:\mathbb{R^2}\to \mathbb{R},\ J_{\omega}|_{\gamma(t)}$ is just the gradient $\nabla \omega|_{\gamma(t)}=(\frac{\partial \omega}{\partial x},\frac{\partial \omega}{\partial y})|_{\gamma(t)}$. For curves $\gamma:\mathbb{R}\to \mathbb{R^2}$ the Jacobian $J_\gamma|_t$ is just the tangent vector $(x'(t),y'(t))$.

Putting all these observations in (1) we end up with \begin{align} \frac{\partial}{\partial t}f(t)&=e^{2\omega(\gamma(t))}\cdot \bigl(\frac{\partial \omega}{\partial x}|_{\gamma(t)},\frac{\partial \omega}{\partial y}|_{\gamma(t)}\bigr)\cdot(x'(t),y'(t))\\ &=e^{2\omega(x,y)}\cdot \bigl(x'(t)\cdot \frac{\partial \omega}{\partial x}+y'(t)\cdot \frac{\partial \omega}{\partial y}\bigr) \end{align}