How to find partial derivatives of $f(x,y)$ with respect to $x$ and $y$

Question

Let $f(x,y) = x^5 \sin (xy^2)$, then find partial derivatives of $f(x,y)$ with respect to $x$ and $y$ at $x=2, y=0.$

Answer 1

I'll do the $y$ partial, then leave the $x$ one to you, as well as evaluating them. To take the $y$ partial derivative, we take the derivative while leaving $x$ fixed (think of it as number like $5$, if it helps you not treat it as a variable), making this a function of just the $y$ variable. Doing so, $$\frac{\partial}{\partial y} \left(x^5\sin (xy^2)\right)=x^5\cos (xy^2) (2xy)=2x^6y\cos (xy^2).$$

Answer 2

[1] Leibniz rule $(fg)_x=f_xg+fg_x$

[2] Chain rule $(f\circ g\ (x))_x=f' g'(x)$

$$f(x,y)=h\sin\ T,\ h(x)=x^5,\ T(x,y)=xy^2$$ so that $$ T(2,0)=0,\ T_x(2,0)=y^2 =0,\ T_y(2,0)=2xy=0$$

Hence $ f_x(2,0) =h_x\sin\ T +h\cos\ T\cdot T_x =0$ and $f_y(2,0) =0$