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Compute partial derivatives of the given function

by RoRi

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:

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

One comment

  1. Mihajlo says:

    There is a typo in the answer for df2/dx2 where it should be 4x^2*y^6 instead of 4x^3*y^6. Also for df2/dy2 it should be 9*x^4*y^4 instead of 6*x^4*y^4 (the book is correct too).

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