Answer to Question #249197 in Calculus for moe

Question

Find the second partial derivative $\displaystyle{\frac{\partial^2 z}{\partial x^2}}$ of the function $z=\sin(xy^2)$ at point $(\pi, 1)$

Answer 1

The first derivative:

$$\dfrac{\partial}{\partial x}(z)=\dfrac{\partial}{\partial x}(\sin(xy^2)) = y^2\cos(xy^2)$$

The second derivative:

$$\dfrac{\partial^2z}{\partial x^2} = \dfrac{\partial}{\partial x}(y^2\cos(xy^2)) = -y^4\sin(xy^2)$$

Substituting $(\pi,1)$, obtain:

$$\dfrac{\partial^2z}{\partial x^2}\vert_{(\pi,1)} = -1^4\cdot \sin(\pi\cdot 1^2) = 0$$

Answer. 0.