Partial Derivative

Partial derivatives are a key concept in multivariable calculus, essential for understanding how functions change with respect to multiple variables. In this blog, we'll dive deep into the concept of partial derivatives, exploring everything from first order to higher-order derivatives, cross partial derivatives, and the rules and formulas that govern them. We'll also discuss how to apply implicit differentiation and integrate partial derivatives, providing examples along the way.

1.0What is a Partial Derivative?

A partial derivative is a derivative where we hold some variables constant and differentiate with respect to one variable in a multivariable function. This concept is crucial in multivariable calculus, where functions depend on more than one variable.

Definition:

Given a function f(x, y, z), which depends on the variables x, y, and z, the partial derivative of f with respect to x (denoted as $\frac{\partial f}{\partial x}$ ) measures the rate of change of f as x changes, while keeping y and z constant.

Example:

Consider the function f(x, y) = x²y + 3xy². The partial derivatives of f with respect to x and y are calculated as follows:

1. Partial derivative with respect to x:

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

Since y is treated as a constant, differentiate x²y and 3xy² with respect to x:

$\frac{\partial f}{\partial x}=2xy+3y^2$

2. Partial derivative with respect to y:

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

Here, treat x as a constant:

$\frac{\partial f}{\partial y}=x^2+6xy$

2.0First-Order Partial Derivatives

First-order partial derivatives represent the rate of change of the function with respect to each variable independently. For a function f(x, y), the first-order partial derivatives are:

$\frac{\partial f}{\partial x}\text{ and }\frac{\partial f}{\partial y}$

These derivatives are crucial in determining the slope of the function in different directions.

3.0Second-Order Partial Derivatives

Second-order partial derivatives provide insights into the curvature of the function. These are the derivatives of the first-order partial derivatives. For a function f(x, y), the second-order partial derivatives are:

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

The term $\frac{{\partial }^{2}f}{\partial x\partial y}$ is known as the cross partial derivative or mixed partial derivative.

4.0Cross Partial Derivatives

Cross partial derivatives involve differentiating a function with respect to one variable and then with respect to another. For a function f(x, y), the cross partial derivatives$\frac{{\partial }^{2}f}{\partial x\partial y}and\frac{{\partial }^{2}f}{\partial y\partial x}$  are generally equal if the function is continuous and well-behaved.

5.0Higher-Order Partial Derivatives

Higher-order partial derivatives extend beyond second order, involving derivatives of derivatives. These are useful in more complex analyses, such as in Taylor series expansions or solving partial differential equations (PDEs).

6.0Partial Derivatives and Implicit Differentiation 

Implicit differentiation is a technique used when a function is defined implicitly rather than explicitly. For example, if F(x, y) = 0 , we might want to find $\frac{dy}{dx}$ using partial derivatives. The chain rule is crucial in implicit differentiation, enabling us to differentiate both sides of an equation with respect to one variable while considering the other as a function.

Example:

Consider x² + y² = 1. To find $\frac{dy}{dx}$ , implicitly differentiate both sides:

$2x+2y\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$

7.0Integrating Partial Derivatives

Integrating partial derivatives involves reversing the differentiation process. If you have a partial derivative and need to find the original function, you'll integrate with respect to the variable. This process is often applied in physics and engineering to determine potential functions from force fields. 

8.0Partial Derivative Equations and Rules

Partial derivative equations (PDEs) are equations that include partial derivatives of an unknown function concerning several variables. They are used to describe various physical phenomena, such as heat conduction, wave propagation, and fluid dynamics.

Key rules for partial derivatives include:

• Sum Rule: The derivative of a sum equals the sum of the derivatives of each term.
• Product Rule: The derivative of a product is determined by $\frac{\partial }{\partial x}(uv)=u\frac{\partial v}{\partial x}+v\frac{\partial u}{\partial x}$.
• Quotient Rule: The derivative of a quotient is determined by $\frac{\partial }{\partial x}\left(\frac{u}{v}\right)=\frac{v\frac{\partial u}{\partial x}-u\frac{\partial v}{\partial x}}{v^2}$.
• Chain Rule: For a function z = f(x, y) where x and y are themselves functions of t, the derivative with respect to t is $\frac{dz}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}.$

Example: 

Let's find the partial derivatives of the function f(x, y) = x²y + sin(xy).

• First-Order Partial Derivatives:

$\frac{\partial f}{\partial x}=2xy+y\cos (xy)$

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

• Second-Order Partial Derivatives:

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

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

9.0Solved Examples on Partial Derivatives 

Example 1: Find $\frac{\partial f}{\partial x}\text{ of }f(x,y)=3x^{2}y+2y^3$

Solution: 

Given: f(x, y) = 3x²y + 2y³

Partial Differentiate with respect to x

⇒ $\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left(3x^{2}y+2y^3\right)$

⇒ $\frac{\partial f}{\partial x}=6xy$

Example 2: Find  $\frac{\partial f}{\partial x}\text{ and }\frac{\partial f}{\partial y}$ of function f(x, y) = x sin y + y cos x.

Solution: 

Given: f(x, y) = x sin y + y cos x.

Partial Differentiate with respect to x

⇒ $\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}(siny+ycosx)$

⇒ $\frac{\partial \mathrm{f}}{\partial \mathrm{x}}$ = sin y – y sin x

⇒ $\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=\frac{\partial }{\partial \mathrm{y}}(xsiny+ycosx)$

⇒$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}$= x cos y + cos x

Example 3: Find $\frac{\partial \mathrm{f}}{\partial \mathrm{x}}$ of function f(x, y) = (x² +y²) ·ln (xy)]

Solution: 

Given: f(x, y) = [(x² +y²) ·ln (xy)]

Partial Differentiate with respect to x

⇒  $\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=\frac{\partial }{\partial \mathrm{x}}[(x2+y2)\cdot In(xy)]$

Using product Rule

⇒ $\frac{\partial f}{\partial x}=\ln (xy)\cdot \frac{\partial }{\partial x}\left(x^2+y^2\right)+\left(x^2+y^2\right)\cdot \frac{\partial }{\partial x}\ln (xy)$

⇒ $\frac{\partial f}{\partial x}=\ln (xy)\cdot 2x+\left(x^2+y^2\right)\cdot \frac{1}{x}$

⇒ $\frac{\partial f}{\partial x}=2x\cdot \ln (xy)+\frac{x^2+y^2}{x}$

Example 4: Find $\frac{\partial \mathrm{y}}{\partial \mathrm{x}}$  of equation x² + xy +y² = 1

Solution: 

Given: x² +xy + y² = 1

Partial differentiation w.r.t. x

⇒ $\frac{\partial }{\partial x}\left(x^2+xy+y^2\right)=\frac{\partial }{\partial x}(1)$

⇒ $2x+y+\frac{x\partial y}{\partial x}+2y\frac{\partial y}{\partial x}=0$

⇒ $\frac{\partial y}{\partial x}=\frac{-2x-y}{x+2y}$

Example 5: If x^x y^y z^z = c. show that at x = y = z $\frac{{\partial }^{2}z}{\partial x\partial y}=-(x\log ex{)}^{-1}$

Solution: 

⇒ log (x^x y^y z^z) = log c [log ab = loga + logb]

⇒ log x^x + logy^y + log z^z = log c [ log a^x = x log a]

⇒ x log x + y logy + zlogz = log c 

Partial differentiation w.r.t x

⇒ $\left(\log x+x\cdot \frac{1}{x}\right)+\left(\log z+z\cdot \frac{1}{z}\right)\frac{\partial z}{\partial x}=0$

⇒ $(1+\log x)+(1+\log z)\frac{\partial z}{\partial x}=0$

⇒ $\frac{\partial z}{\partial x}=-\frac{(1+\log x)}{(1+\log z)}$

Similarly, $\frac{\partial z}{\partial y}=-\frac{(1+\log y)}{(1+\log z)}$

Partial differentiation w.r.t. x

⇒ $\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right)=\frac{-\partial }{\partial x}\left(\frac{1+\log y}{1+\log z}\right)$

⇒ $\frac{{\partial }^{2}z}{\partial x\partial y}=-(1+\log y)\left[\frac{-1}{(1+\log z{)}^2}\cdot \frac{1}{z}\cdot \frac{\partial z}{\partial x}\right]$

⇒ $\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{(1+\log y)}{(1+\log z{)}^2}\cdot \frac{1}{z}\cdot \frac{(1+\log x)}{(1+\log z)}$

⇒ ${\left(\frac{{\partial }^{2}z}{\partial x\partial y}\right)}_{x=y=z}=-\frac{(1+\log x{)}^2}{x(1+\log x{)}^3}$

⇒\left(\frac{\partial^2z}{\partialx\partialy}\right)_{x=y=z}=\frac{-1}{x(1+\log x)}

⇒ \left(\frac{\partial^2z}{\partialx \partial y}\right)_{x=y=z}=\frac{-1}{x\left(\log _e e+\log _e x\right)}   

⇒  ${\left(\frac{{\partial }^{2}z}{\partial x\partial y}\right)}_{x=y=z}=\frac{1}{x[\log (ex)]}$

⇒  ${\left(\frac{{\partial }^{2}z}{\partial x\partial y}\right)}_{x=y=z}=(x\log ex{)}^{-1}$

Example 6: If u = exyz . Prove that $\frac{{\partial }^{3}u}{\partial x\partial y\partial z}=\left(1+3xyz+x^2{y}^2{z}^2\right)e^{xyz}$

Solution: 

Given u = e^xyz 

then $\frac{\partial \mathrm{u}}{\partial \mathrm{z}}={\mathrm{xye}}^{\mathrm{xyz}}$

 Now partial derivate with respect to y.

⇒ $\frac{\partial }{\partial \mathrm{y}}\left(\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right)=\frac{\partial }{\partial \mathrm{y}}\left({\mathrm{xye}}^{\mathrm{xyz}}\right)$

⇒ $\frac{{\partial }^{2}u}{\partial y^{\partial }z}=xy\frac{\partial }{\partial y}\left(e^{xyz}\right)+e^{xyz}\frac{\partial }{\partial y}(xy)$

⇒ $\frac{{\partial }^{2}u}{\partial y\partial z}=xy(xz)e^{xyz}+x\cdot e^{xyz}$

⇒ $\frac{{\partial }^{2}u}{\partial y\partial z}=e^{xyz}\left(x^{2}yz+x\right)$

Now partial derivate with respect to x.

⇒ $\frac{\partial }{\partial x}\left(\frac{{\partial }^{2}u}{\partial y\partial z}\right)=\frac{\partial }{\partial x}\left[\left(x^{2}yz+x\right)e^{xyz}\right]$

⇒ $\frac{{\partial }^{3}y}{\partial x\partial y\partial z}=\left(x^{2}yz+x\right)\frac{\partial }{\partial x}\left(e^{xyz}\right)+e^{xyz}\frac{\partial }{\partial x}\left(x^{2}yz+x\right)$

⇒ $\frac{{\partial }^{3}y}{\partial x\partial y\partial z}=\left(x^{2}zy+x\right)\left(yz{e}^{xyz}\right)+e^{xyz}(2xyz+1)$

⇒ $\frac{{\partial }^{3}y}{\partial x\partial y\partial z}=\left(x^2{y}^2{z}^2+xyz\right)e^{xyz}+e^{xyz}(2xyz+1)$

⇒ $\frac{{\partial }^{3}y}{\partial x\partial y\partial z}=\left(x^2{y}^2{z}^2+xyz+2xyz+1\right)e^{xyz}$

⇒ $\frac{{\partial }^{3}y}{\partial x\partial y\partial z}=\left(1+3xyz+x^2{y}^2{z}^2\right)e^{xyz}$

10.0Practice Questions on Partial Derivatives

1. Find $\frac{\partial f}{\partial x}\text{ and }\frac{\partial f}{\partial y}$ of function f(x, y) = 5x²y + 3xy² 
2. Find $\frac{{\partial }^{2}f}{\partial x^2},\frac{{\partial }^{2}f}{\partial y^2},\frac{{\partial }^{2}f}{\partial x\partial y}$ of function f(x, y) = x²y² + x²y² + e^xy 
3. Given the equation x²+ y² +z² = 1, Find $\frac{\partial z}{\partial x}\text{ and }\frac{\partial z}{\partial y}$
4. Given the function f(x, y) = x²y² + sin (xy) compute $\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}\text{ and }\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{y}\partial \mathrm{x}}$
5. If z (x + y) = x² + y². Show that ${\left(\frac{\partial \mathrm{z}}{\partial \mathrm{x}}-\frac{\partial \mathrm{z}}{\partial \mathrm{y}}\right)}^2=4\left(1-\frac{\partial \mathrm{z}}{\partial \mathrm{x}}-\frac{\partial \mathrm{z}}{\partial \mathrm{y}}\right)$.

11.0Sample Questions on Partial Derivatives

Q. What are higher-order partial derivatives?

Ans: Higher-order partial derivatives refer to the derivatives of partial derivatives. For example, if f(x, y) is a function, the first-order partial derivatives are $\frac{\partial f}{\partial x}and\frac{\partial f}{\partial y}$ . The second-order partial derivatives include $\frac{{\partial }^{2}f}{\partial x^2},\frac{{\partial }^{2}f}{\partial y^2}$ , and mixed derivatives like $\frac{{\partial }^{2}f}{\partial x\partial y}$.

Q. What are mixed partial derivatives?

Ans: Mixed partial derivatives are second or higher-order derivatives where differentiation is performed with respect to different variables. For instance, for a function f(x, y), $\frac{{\partial }^{2}f}{\partial x\partial y}$ is a mixed partial derivative. If f(x, y) is sufficiently smooth, then $\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{{\partial }^{2}f}{\partial y\partial x}$.