# Math Help - mixed partial derivatives

1. ## mixed partial derivatives

f = exp(-xy^2)

find $\frac{\partial^2 f}{\partial x \partial y}$

and $\frac{\partial^2 f}{\partial y \partial x}$

2. Originally Posted by sonia1

f = exp(xy^2)

find $\frac{\partial^2 f}{\partial x \partial y}$

and $\frac{\partial^2 f}{\partial y \partial x}$

For $\frac{\partial^2 f}{\partial x \partial y}$, differentiate with respect to x first, then with respect to y. Remember, when you differentiate with respect x, you pretend all other variables are constants. And when you differentiate with respect to y, you pretend all other variables are constants.

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \exp(xy^2)$

By the chain rule:

$\frac{\partial f}{\partial x} = \exp(xy^2) \times \frac{\partial}{\partial x} (xy^2)$

$\frac{\partial f}{\partial x} = \exp(xy^2) \times y^2$

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial y} (\frac{\partial f}{\partial x}) = \frac{\partial }{\partial y} ( \exp(xy^2) \times y^2)$

By the product rule:

$\frac{\partial^2 f}{\partial x \partial y} = \exp(xy^2) \frac{\partial}{\partial y} (y^2) + y^2 \times \frac{\partial }{\partial y} \exp(xy^2)$

Now carry out the differentiation, remembering to use the product rule again on the 2nd term.

$\frac{\partial^2 f}{\partial x \partial y} =\exp(xy^2) \times 2y + y^2 \exp(xy^2) \times \frac{\partial }{\partial y} (xy^2)$

$\frac{\partial^2 f}{\partial x \partial y} =\exp(xy^2) \times 2y + y^2 \exp(xy^2) \times 2xy$

$\frac{\partial^2 f}{\partial x \partial y} =\exp(xy^2) \times 2y +2xy^3 \exp(xy^2)$

Now do the same again, only change the order of differentiation. Wrt y first, then wrt x. See if you get the same result.

In general $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$ if the function is sufficiently smooth.

3. Hello, sonia1!

Watch out for "products" . . .

$f(x,y) \:=\:e^{xy^2}$

Find: . $\frac{\partial^2 f}{\partial x \partial y}\,\text{ and }\,\frac{\partial^2 f}{\partial y \partial x}$

Are they equal? . . . . yes

$\frac{\partial f}{\partial y} \;=\;e^{xy^2}\!\cdot\!2xy \;=\;2xy\!\cdot\!e^{xy^2}$

$\frac{\partial^2y}{\partial x\partial y} \;=\;2xy\!\cdot\!e^{xy^2}\!\cdot\!y^2 + 2y\!\cdot\! e^{xy^2} \;=\;2ye^{xy^2}(xy^2+1)$

$\frac{\partial f}{\partial x} \;=\;e^{xy^2}\!\cdot\!y^2 \;=\;y^2\!\cdot\!e^{xy^2}$

$\frac{\partial^2f}{\partial y\partial x} \;=\;y^2\!\cdot\!e^{xy^2}\!\cdot\!2xy + 2y\!\cdot\!e^{xy^2} \;=\;2ye^{xy^2}(xy^2+1)$

4. Originally Posted by Mush
For $\frac{\partial^2 f}{\partial x \partial y}$, differentiate with respect to x first, then with respect to y. Remember, when you differentiate with respect x, you pretend all other variables are constants. And when you differentiate with respect to y, you pretend all other variables are constants.

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \exp(xy^2)$

By the chain rule:

$\frac{\partial f}{\partial x} = \exp(xy^2) \times \frac{\partial}{\partial x} (xy^2)$

$\frac{\partial f}{\partial x} = \exp(xy^2) \times y^2$

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial y} (\frac{\partial f}{\partial x}) = \frac{\partial }{\partial y} ( \exp(xy^2) \times y^2)$

By the product rule:

$\frac{\partial^2 f}{\partial x \partial y} = \exp(xy^2) \frac{\partial}{\partial y} (y^2) + y^2 \times \frac{\partial }{\partial y} \exp(xy^2)$

Now carry out the differentiation, remembering to use the product rule again on the 2nd term.

$\frac{\partial^2 f}{\partial x \partial y} =\exp(xy^2) \times 2y + y^2 \exp(xy^2) \times \frac{\partial }{\partial y} (xy^2)$

$\frac{\partial^2 f}{\partial x \partial y} =\exp(xy^2) \times 2y + y^2 \exp(xy^2) \times 2xy$

$\frac{\partial^2 f}{\partial x \partial y} =\exp(xy^2) \times 2y +2xy^3 \exp(xy^2)$

Now do the same again, only change the order of differentiation. Wrt y first, then wrt x. See if you get the same result.

In general $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$ if the function is sufficiently smooth.
what do you mean by it is equal if the fuction is sufficiently smooth

5. Originally Posted by Soroban
Hello, sonia1!

Watch out for "products" . . .

$\frac{\partial f}{\partial y} \;=\;e^{xy^2}\!\cdot\!2xy \;=\;2xy\!\cdot\!e^{xy^2}$

$\frac{\partial^2y}{\partial x\partial y} \;=\;2xy\!\cdot\!e^{xy^2}\!\cdot\!y^2 + 2y\!\cdot\! e^{xy^2} \;=\;2ye^{xy^2}(xy^2+1)$

$\frac{\partial f}{\partial x} \;=\;e^{xy^2}\!\cdot\!y^2 \;=\;y^2\!\cdot\!e^{xy^2}$

$\frac{\partial^2f}{\partial y\partial x} \;=\;y^2\!\cdot\!e^{xy^2}\!\cdot\!2xy + 2y\!\cdot\!e^{xy^2} \;=\;2ye^{xy^2}(xy^2+1)$
sorry i wrote the function wrong.

Also, what do u get when f = cos(y/x)?

6. ## thankyou

hey thankx a lot guyz I understand partial derivatives now