Let $g:\mathbb{R}^{2} \rightarrow \mathbb{R}$, such that $g(x,y)=f(\cos(xy),y^3-x^2)$. We know the second-order Taylor polynomial of $f$ centered on the point $(1,-1)$ is $T(x,y)=2-2x+y-3x^2+xy+4y^2$.
I'm asked to calculate the directional derivative of $g$ in the direction of the vector $v=(-1,3)$ on the point $(0,-1)$, given the data above.
Let's say I name $h(x,y)=(\cos(xy),y^3-x^2)$. Now, I know that I can just apply the chain rule to $f$ on $(0,-1)$. That gives me $\nabla g(0,-1) = \nabla f(h(0,-1))\cdot D_h(0,-1) $ , and then I solve it considering the value of the Taylor polynomial at $(1,-1)$.
But I was thinking if it's possible to simply state that $\nabla g(0,1)= \nabla f(1,-1) $ at the given point $(0,1)$ using the partial derivatives of the Taylor polynomial.
Thanks a lot in advance!
It holds $g(\mathbf{x}) = f(\mathbf{u})$ Chain rule indicates the gradient $$ \frac{\partial g}{\partial \mathbf{x}} (\mathbf{x}_0) = \mathbf{J}^T(\mathbf{x}_0) \frac{\partial f}{\partial \mathbf{u}}(\mathbf{u}_0) $$
Here $\mathbf{x}_0=(0,-1)$ and $\mathbf{u}_0=(1,-1)$ and $\mathbf{J}$ is the Jacobian of the transformation, aka a 2-by-2 matrix.
Because you are given the Taylor approximation of $f$ at point $\mathbf{u}_0$, you can easily compute the gradient on the RHS.
Finally you are asked to compute the directional derivative as $\frac{\partial g}{\partial \mathbf{x}} (\mathbf{x}_0) : \mathbf{v} $ where the colon denotes the dot product between two vectors. Can you finish from here ?