Find $\nabla F (2,0,-3)$ given the data

Question

Two functions are composed so $ F(u,v,w)=f(x(u,v,w),y(u,v,w))$ and it is known that $$x(-2,0,3)=-1 \,\,\,\,\,\,, \,\,\,\,\,\,y(-2,0,3)=1$$ $$\nabla x(-2,0,3)=(\frac{1}{2},\frac{1}{3},\frac{1}{4}) \,\,\,\,\,\,, \,\,\,\,\,\,\nabla y(-2,0,3)=(\frac{1}{5},\frac{1}{6},\frac{1}{7}) $$ $$\nabla f(-1,1)=(\sqrt{2}, \sqrt{3} ) $$
Aslo it is known that $x$ and $y$ are differentiable on $(-2,0,3)$ and that $f$ is differentiable on $(-1,1)$ Given this I can find that $F(-2,0,3)=(\sqrt{2}, \sqrt{3} )$ but can I find $\nabla F (-2,0,3)$ given the data that I have? and if so how?

Answer 1

Writing: $$\vec{x} = \langle x,y \rangle \qquad \vec{u} = \langle u,v,w \rangle $$ we find the derivative matrix using the chain rule: \begin{align*} F(\vec{u}) &= f(\vec{x}(\vec{u})) \\ \mathbf{D}F(\vec{u}) &= \mathbf{D}f(\vec{x}(\vec{u})) \cdot \mathbf{D}\vec{x}(\vec{u}) \\ \mathbf{D}F(\vec{u})|_{\vec{u}=(-2,0,3)} &= \mathbf{D}f(\vec{x}(\vec{u}))|_{\vec{u}=(-2,0,3)} \cdot \mathbf{D}\vec{x}(\vec{u})|_{\vec{u}=(-2,0,3)} \\ &= \begin{bmatrix} \sqrt{2} & \sqrt{3} \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{5} & \frac{1}{6} & \frac{1}{7} \end{bmatrix} \\ &= \begin{bmatrix} \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{5} & \frac{\sqrt{2}}{3} + \frac{\sqrt{3}}{6} & \frac{\sqrt{2}}{4} + \frac{\sqrt{3}}{7} \end{bmatrix} \end{align*}

More often, rather than using the derivative matrices and chain rule in the above form, it is commonly seen to write out the equations: \begin{align*} \frac{\partial F}{\partial u} &= \frac{\partial}{\partial u}f(x(u,v,w),y(u,v,w)) = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u} \\ \frac{\partial F}{\partial v} &= \frac{\partial}{\partial v}f(x(u,v,w),y(u,v,w)) = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} \\ \frac{\partial F}{\partial w} &= \frac{\partial}{\partial w}f(x(u,v,w),y(u,v,w)) = \frac{\partial f}{\partial x}\frac{\partial x}{\partial w} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial w} \end{align*}

which arise from seeing the relations between the variables:

Both ways give the same answer and are the same concept - the matrix form of the chain rule is simply a way to generalize and shorthand. The derivative (the gradient) evaluated at the point is then:

$$ \mathbf{D} F(-2,0,3) = \begin{bmatrix} \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{5} & \frac{\sqrt{2}}{3} + \frac{\sqrt{3}}{6} & \frac{\sqrt{2}}{4} + \frac{\sqrt{3}}{7} \end{bmatrix} $$