Calculating gradient, how to do this?

Question

Given $f(x,y)$, a function that has continuous partial derivatives in every point.

such that $\nabla f(0,-18)=-2i+3j$

We define a new function $g(x,y)=f(xy+x^2,xy-y^2)$ calculate $\nabla g(3,-3)$

How I tried to solve this? I need to find: $$\nabla g(3,-3) = g_x'(3,-3)i+g_y'(3,-3)j=f(xy+x^2,xy-y^2)_x'(3,-3)i+f(xy+x^2,xy-y^2)_y'(3,-3)j$$

and I got stuck here; I don't have f to calculate the partial directive for it...

Answer 1

To simplify notation, let us define $u= xy+x^2$, $v=xy-y^2$, so that: $$ \frac{\partial u}{\partial x}=y+2x ,\quad \frac{\partial v}{\partial x}=y $$

Then, using the chain rule, since $g(x,y)=f(u,v)$: $$\frac{\partial g}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}$$

Now, if $q=(x,y)=(3,-3)$ we have $p=(u,v)=(0,-18)$, and we already know: $$ \left.\frac{\partial f}{\partial u}\right|_p =-2 , \quad \left.\frac{\partial f}{\partial v}\right|_p =3 $$ Also we can compute from above: $$ \left.\frac{\partial u}{\partial x}\right|_q =3 ,\quad \left. \frac{\partial v}{\partial x}\right|_q=-3 $$

Then, $$\left.\frac{\partial g}{\partial x}\right|_p=-2 \cdot 3+ 3 \cdot (-3)=-15 $$

Can you repeat the operation with $\dfrac{\partial g}{\partial y}$ ?

Answer 2

We can make use of the multivariable chain rule (see this answer) which states that given a function $h: \mathbb{R}^n \to \mathbb{R}^m$ and a function $f: \mathbb{R}^m \to \mathbb{R}$ we have $$ \nabla (f \circ h) = \left(h'\right)^T \cdot \left(\nabla f \circ h\right) $$ where "$\cdot$" represents matrix multiplication.

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In our case, we see that if we define $h(x,y) = (xy+x^2,xy-y^2)$ then it follows from the definition of $g$ that $$g(x,y) = (f \circ h)(x,y)= f(xy+x^2,xy-y^2) \tag{1}$$ which means that calculating $\nabla g(3,-3)$ is equal to calculating $\nabla (f \circ h)(3,-3) $.

Now, by definition, we know that $$ h' = \begin{pmatrix} \frac{\partial}{\partial x}(xy+x^2) & \frac{\partial}{\partial y}(xy+x^2)\\ \frac{\partial}{\partial x}(xy-y^2) & \frac{\partial}{\partial y}(xy-y^2) \end{pmatrix} = \begin{pmatrix} y+2x & x\\ y & x-2y \end{pmatrix} $$ which means that the transposed matrix $\left(h'\right)^T$ is $\begin{pmatrix} y+2x & y\\ x & x-2y \end{pmatrix} $. If we then evaluate this matrix at $(3,-3)$ we get $$ \left(h'\right)^T(3,-3) = \begin{pmatrix} -3+2(3) & -3\\ 3 & 3-2(-3) \end{pmatrix} = \begin{pmatrix} -3+6 & -3\\ 3 & 3+6 \end{pmatrix} = \begin{pmatrix} 3 & -3\\ 3 & 9 \end{pmatrix} $$ On the other hand, we see that $$ h(3,-3) = (3(-3)+3^2, 3(-3) -(-3)^2) = (-9+9, -9-9) = (0,-18) $$ which tells us that $$ \left(\nabla f \circ h\right)(3,-3) = \nabla f\left(h(3,-3)\right) = \nabla f\left(0,-18\right) = \begin{pmatrix} -2 \\3 \end{pmatrix} $$ using the convention of the gradient as a column vector.

Finally, putting all this together tells us that $$ \nabla g(3,-3) = \nabla (f \circ h)(3,-3) = \left[\left(h'\right)^T(3,-3)\right] \cdot \left[\left(\nabla f \circ h\right)(3,-3)\right] = \begin{pmatrix} 3 & -3\\ 3 & 9 \end{pmatrix} \cdot \begin{pmatrix} -2 \\3 \end{pmatrix} = \begin{pmatrix} -2(3)+3(-3) \\-2(3)+3(9) \end{pmatrix}= \begin{pmatrix} -6-9 \\-6+27 \end{pmatrix}= \begin{pmatrix} -15 \\21 \end{pmatrix} = -15 \boldsymbol{\hat\imath} + 21 \boldsymbol{\hat\jmath} $$