Home » Blog » Compute partial derivatives of the given function

Compute partial derivatives of the given function

Let

    \[ f(x,y) = \sin(x^2 y^3). \]

Compute all first-order and second-order partial derivatives. Verify that the mixed partial derivatives are equal.


The first-order partial derivatives are given by

    \begin{align*}  \frac{\partial f}{\partial x} &= 2xy^3 \cos (x^2 y^3); \\  \frac{\partial f}{\partial y} &= 3x^2 y^2 \cos (x^2 y^3). \end{align*}

The second partial derivatives are given by

    \begin{align*}  \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \\   &= \frac{\partial}{\partial x} (2xy^3 \cos (x^2 y^3))  \\  &= 2y^3 \cos (x^2 y^3) - 4x^3y^6 \sin (x^2 y^3). \\  \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\  &= \frac{\partial}{\partial y} (2xy^3 \cos (x^2 y^3)) \\  &= 6xy^2 \cos (x^2 y^3) - 6x^3 y^5 \sin (x^2 y^3).\\  \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \\  &= \frac{\partial}{\partial x} (3x^2 y^2 \cos (x^2 y^3)) \\  &= 6xy^2 \cos (x^2 y^3) - 6x^3 y^5 \sin (x^2 y^3).\\  \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\  &= \frac{\partial}{\partial y} (3x^2 y^2 \cos (x^2 y^3)) \\  &= 6x^2 y \cos (x^2 y^3) - 6x^4 y^4 \sin (x^2 y^3). \end{align*}

From this we see that the mixed partials are equal:

    \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 6xy^2 - \cos(x^2 y^3) - 6x^3 y^5 \sin (x^2 y^3). \]

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):