Tag: Partial Derivatives

Prove formulas for the partial derivatives of x^y

by RoRi

November 17, 2015

Define a function of two variables, with $x> 0$ ,

$f(x,y) = x^y.$

Prove

$\frac{\partial f}{\partial x} = yx^{y-1} \qquad \text{and} \qquad \frac{\partial f}{\partial y} = x^y \log x.$

Proof. First, we write,

$f(x,y) = x^y = e^{ y \log x}.$

Then we compute the derivatives using the chain rule and formulas for the derivatives of the exponential and logarithm,

$\frac{\partial f}{\partial x} = e^{y \log x} \cdot \frac{y}{x} = yx^{y-1}.$

And,

$\frac{\partial f}{\partial y} = e^{y \log x} \cdot \log x = x^y \log x. \qquad \blacksquare$

Show a given equation satisfies a particular second or partial differential equation

by RoRi

October 14, 2015

Let

$f(x,y) = \frac{xy}{(x^2+y^2)^2}, \qquad (x,y) \neq (0,0).$

Show that

$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0.$

Proof. First, we compute the first-order partial derivative with respect to $x$ ,

$\begin{align*} \frac{\partial f}{\partial x} &= \frac{y(x^2+y^2)^2 - (xy)(2)(2x)(x^2+y^2)}{(x^2+y^2)^4} \\ &= \frac{x^2y + y^3 - 4x^2y}{(x^2+y^2)^3} \\ &= \frac{y^3 - 3x^2y}{(x^2+y^2)^3}. \end{align*}$

Then, we compute the first-order partial derivative with respect to $y$ ,

$\begin{align*} \frac{\partial f}{\partial y} &= \frac{x(x^2+y^2)^2 - (xy)(2)(2y)(x^2+y^2)}{(x^2+y^2)^4} \\ &= \frac{x^3 + xy^2 - 4xy^2}{(x^2+y^2)^3} \\ &= \frac{x^3 - 3xy^2}{(x^2+y^2)^3}. \end{align*}$

Next, we compute the second-order partial derivatives in the statement we are trying to prove,

$\begin{align*} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{y^3 - 3x^2y}{(x^2+y^2)^3} \right) \\ &= \frac{-6xy (x^2+y^2)^3 - (y^3-3x^2y)(3)(2x)(x^2+y^2)^2}{(x^2+y^2)^6} \\ &= \frac{-6x^3y - 6xy^3 - 6xy^3 + 18x^3y}{(x^2+y^2)^4} \\ &= 12xy \frac{x^2 - y^2}{(x^2+y^2)^4}. \\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{x^3 - 3xy^2}{(x^2+y^2)^3} \right) \\ &= \frac{-6xy(x^2+y^2)^3 - (x^3 - 3xy^2)(3)(2y)(x^2+y^2)^2}{(x^2+y^2)^6} \\ &= \frac{-6x^3 y - 6xy^3 - 6x^3 y + 18xy^3}{(x^2+y^2)^4} \\ &= -12xy \frac{x^2 - y^2}{(x^2+y^2)}. \end{align*}$

Finally, putting these together we have,

$\begin{align*} \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} &= 12xy \frac{x^2 - y^2}{(x^2+y^2)^4} - 12xy \frac{x^2 - y^2}{(x^2+y^2)^4} \\ &= 0. \qquad \blacksquare \end{align*}$

Prove two formulas with partial derivatives

by RoRi

October 14, 2015

Show that

$x \left( \frac{\partial z}{\partial x} \right) + y \left( \frac{\partial z}{\partial y} \right) = 2z$

when

$z = (x-2y)^2$ ;
$z = (x^4 + y^4)^{\frac{1}{2}}$ .

We compute,
$\begin{align*} x \left( \frac{\partial z}{\partial x} \right) + y \left( \frac{\partial z}{\partial y} \right) &= x \left( \frac{\partial}{\partial x} (x-2y)^2 \right) + y \left( \frac{\partial}{\partial y} (x-2y)^2 \right) \\ &= x (2 (x-2y)) + y (2(-2)(x-2y)) \\ &= 2x^2 - 4xy - 4xy + 8y^2 \\ &= 2 (x^2 - 4xy + 4y) \\ &= 2 (x - 2y)^2 \\ &= 2z. \end{align*}$
We compute,
$\begin{align*} x \left( \frac{\partial z}{\partial x} \right) + y \left( \frac{\partial z}{\partial y} \right) &= x \left( \frac{\partial}{\partial x} (x^4 + y^4)^{\frac{1}{2}} \right) + y \left( \frac{\partial}{\partial y} (x^4+y^4)^{\frac{1}{2}} \right) \\ &= x (4x^3)(1/2)(x^4+y^4)^{-\frac{1}{2}} + y (4y^3) (1/2) (x^4+y^4)^{-\frac{1}{2}} \\ &= \frac{2x^4 + 2y^4}{\sqrt{x^4+y^4}} \\ &= 2 (x^4+y^4)^{\frac{1}{2}} \\ &= 2z. \end{align*}$

Compute partial derivatives of the given function

by RoRi

October 14, 2015

Let

$f(x,y) = \frac{x}{\sqrt{x^2+y^2}} \qquad (x,y) \neq (0,0).$

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

$\begin{align*} \frac{\partial f}{\partial x} &= \frac{\sqrt{x^2+y^2} - \frac{x^2}{\sqrt{x^2+y^2}}}{x^2+y^2} \\ &= \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}}.\\ \frac{\partial f}{\partial y} &= \frac{-x \frac{y}{\sqrt{x^2+y^2}}}{x^2+y^2} \\ &= \frac{-xy}{(x^2+y^2)^{\frac{3}{2}}}. \end{align*}$

The second partial derivatives are given by

$\begin{align*} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial x} \left( \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}} \right) \\ &= \frac{-y^2 (3/2)(2x)(\sqrt{x^2+y^2})}{(x^2+y^2)^3} \\ &= \frac{-3xy^2}{(x^2+y^2)^{\frac{5}{2}}}.\\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial y} \left(\frac{y^2}{(x^2+y^2)^{\frac{3}{2}}} \right) \\ &= \frac{2y(x^2+y^2)^{\frac{3}{2}} - y^2 (3/2)(2y)\sqrt{x^2+y^2}}{(x^2+y^2)^3} \\ &= \frac{2y(x^2+y^2) - 3y^3}{(x^2+y^2)^{\frac{5}{2}}} \\ &= \frac{y(2x^2 - y^2)}{(x^2+y^2)^{\frac{5}{2}}}. \end{align*}$

Then,

$\begin{align*} \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial x} \left( \frac{-xy}{(x^2+y^2)^{\frac{3}{2}}} \right) \\ &= \frac{-y(x^2+y^2)^{\frac{3}{2}} + xy(2x)(3/2)\sqrt{x^2+y^2}}{(x^2+y^2)^3} \\ &= \frac{-yx^2 - y^3 + 3x^2y}{(x^2+y^2)^{\frac{5}{2}}} \\ &= \frac{y (2x^2 y - y^2)}{(x^2+y^2)^{\frac{5}{2}}}.\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} \left(\frac{-xy}{(x^2+y^2)^{\frac{3}{2}}}\right) \\ &= \frac{-x(x^2+y^2)^{\frac{3}{2}} + xy(2y)(3/2) \sqrt{x^2+y^2}}{(x^2+y^2)^3}\\ &= \frac{-x^3 -xy^2 + 3xy^2}{(x^2+y^2)^{\frac{5}{2}}} \\ &= \frac{x (2xy^2 - x^2)}{(x^2+y^2)^{\frac{5}{2}}}. \end{align*}$

From this we see that the mixed partials are equal:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = \frac{y (2x^2 y - y^2)}{(x^2+y^2)^{\frac{5}{2}}}.$

Compute partial derivatives of the given function

by RoRi

October 14, 2015

Let

$f(x,y) = \frac{x+y}{x-y} \qquad (x \neq y).$

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

$\begin{align*} \frac{\partial f}{\partial x} &= \frac{1}{x-y} - \frac{x+y}{(x-y)^2} = \frac{-2y}{(x-y)^2}; \\ \frac{\partial f}{\partial y} &= \frac{1}{x-y} + \frac{x+y}{(x-y)^2} = \frac{2x}{(x-y)^2}. \end{align*}$

The second partial derivatives are given by

$\begin{align*} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial x} \left( \frac{-2y}{(x-y)^2} \right) \\ &= \frac{(-2y)(2)(x-y)}{(x-y)^4} \\ &= \frac{4y}{(x-y)^3}. \\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial y} \left(\frac{-2y}{(x-y)^2} \right) \\ &= \frac{2(x-y)^2 + 2y(2)(x-y)}{(x-y)^4}\\ &= \frac{-2 (x+y)}{(x-y)^3}. \\ \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial x} \left( \frac{2x}{(x-y)^2} \right) \\ &= \frac{2(x-y)^2 - (2x)(2)(x-y)}{(x-y)^4} \\ &= \frac{-2(x+y)}{(x-y)^3}.\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} \left(\frac{2x}{(x-y)^2}\right) \\ &= \frac{(2x)(2)(x-y)}{(x-y)^4} \\ &= \frac{4x}{(x-y)^3}. \end{align*}$

From this we see that the mixed partials are equal:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = \frac{-2 (x+y)}{(x-y)^3}.$

Compute partial derivatives of the given function

by RoRi

October 14, 2015

Let

$f(x,y) = \sin (\cos (2x-3y)).$

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

Using the chain rule, the first-order partial derivatives are given by

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

The second partial derivatives are given by

$\begin{align*} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial x} (-2 \cos (\cos (2x-3y)) \sin (2x-3y)) \\ &= -4(\cos(2x-3y) \cos (\cos (2x-3y)) + (\sin (2x-3y))^2 \sin (\cos (2x-3y))). \\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial y} (-2 \cos (\cos (2x-3y)) \sin (2x-3y)) \\ &= 6\cos (2x-3y) \cos (\cos (2x-3y)) + 6 (\sin (2x-3y))^2 \sin (\cos (2x-3y)).\\ \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial x} (3 \cos (\cos (2x-3y)) \sin (2x-3y)) \\ &= 6 \cos (2x-3y) \cos (\cos (2x-3y)) + 6(\sin (2x-3y))^2 \sin (\cos (2x-3y)).\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} (3 \cos (\cos (2x-3y)) \sin (2x-3y)) \\ &= -9(\cos(2x-3y)\cos(\cos (2x-3y)) + (\sin (2x-3y))^2 \sin (\cos (2x-3y))). \end{align*}$

From this we see that the mixed partials are equal:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 6 \cos (2x-3y) \cos (\cos (2x-3y) + 6(\sin (2x-3y))^2 \sin (\cos (2x-3y)).$

Compute partial derivatives of the given function

by RoRi

October 14, 2015

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:

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

Compute partial derivatives of the given function

by RoRi

October 14, 2015

Let

$f(x,y) = \sqrt{x^2 + y^2}.$

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

$\begin{align*} \frac{\partial f}{\partial x} &= \left( \frac{1}{2} \right) \left( \frac{1}{\sqrt{x^2+y^2}} \right) (2x) = \frac{x}{\sqrt{x^2+y^2}}; \\ \frac{\partial f}{\partial y} &= \left( \frac{1}{2} \right) \left( \frac{1}{\sqrt{x^2+y^2}} \right) (2y) = \frac{y}{\sqrt{x^2+y^2}}. \end{align*}$

The second partial derivatives are given by

$\begin{align*} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial x} \left( \frac{x}{\sqrt{x^2+y^2}} \right) \\ &= \frac{\sqrt{x^2+y^2} - \frac{x^2}{\sqrt{x^2+y^2}}}{x^2+y^2} \\ &= \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}}. \\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial y} \left( \frac{x}{\sqrt{x^2+y^2}} \right) \\ &= \frac{-xy}{(x^2+y^2)^{\frac{3}{2}}}.\\ \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial x} \left(\frac{y}{\sqrt{x^2+y^2}}\right) \\ &= \frac{-xy}{(x^2+y^2)^{\frac{3}{2}}}.\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} \left(\frac{y}{\sqrt{x^2+y^2}}\right) \\ &= \frac{x^2}{(x^2+y^2)^{\frac{3}{2}}}. \end{align*}$

From this we see that the mixed partials are equal:

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

Compute partial derivatives of the given function

by RoRi

October 13, 2015

Let

$f(x,y) = xy + \frac{x}{y} \qquad (y \neq 0).$

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

$\begin{align*} \frac{\partial f}{\partial x} &= y + \frac{1}{y}; \\ \frac{\partial f}{\partial y} &= x - \frac{x}{y^2}. \end{align*}$

The second partial derivatives are given by

$\begin{align*} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial x} \left( y + \frac{1}{y} \right) \\ &= 0. \\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial y} \left(y + \frac{1}{y} \right) \\ &= 1 - \frac{1}{y^2}.\\ \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial x} \left( x - \frac{x}{y^2} \right) \\ &= 1 - \frac{1}{y^2}.\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} \left( x - \frac{x}{y^2} \right) \\ &= \frac{2x}{y^3}. \end{align*}$

From this we see that the mixed partials are equal:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 1 - \frac{1}{y^2}.$

Compute partial derivatives of the given function

by RoRi

October 13, 2015

Let

$f(x,y) = x \sin (x+y).$

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

$\begin{align*} \frac{\partial f}{\partial x} &= \sin (x+y) + x \cos (x+y); \\ \frac{\partial f}{\partial y} &= x \cos (x+y). \end{align*}$

The second partial derivatives are given by

$\begin{align*} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial x} (\sin (x+y) + x \cos (x+y)) \\ &= \cos (x+y) + \cos (x+y) - x \sin (x+y) \\ &= 2 \cos (x+y) - x \sin (x+y). \\ \frac{\partial^2 f}{\partial y \partial x} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ &= \frac{\partial}{\partial y} (\sin (x+y) + x \cos (x+y)) \\ &= \cos(x+y) - x \sin (x+y).\\ \frac{\partial^2 f}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial x} (x \cos (x+y)) \\ &= \cos (x+y) - x \sin (x+y).\\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \\ &= \frac{\partial}{\partial y} (x \cos (x+y)) \\ &= -x \sin (x+y). \end{align*}$

From this we see that the mixed partials are equal:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = \cos(x+y) - x \sin(x+y).$