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# Basics of Partial Differentiation

In Mathematics, sometimes the function depends on two or more than two variables. In this case, the derivative converts into the partial derivative since the function depends on several variables. Whereas, the partial differential equation is an equation containing one or more partial derivatives. Partial derivatives are usually used in vector calculus and differential geometry. In this article, students will learn the basics of partial differentiation.

## Partial Derivative Rules

Just like ordinary derivatives, partial derivatives follow some rules like product rule, quotient rule, chain rule etc.

If u = f(x,y), then the partial derivatives follow some rules as the ordinary derivatives.

Product Rule:

If u = f(x,y).g(x,y), then

$$\begin{array}{l}u_{x}=\frac{\partial u}{\partial x}= g(x, y)\frac{\partial f}{\partial x}+f(x, y)\frac{\partial g}{\partial x}\end{array}$$

And,

$$\begin{array}{l}u_{y}=\frac{\partial u}{\partial y}=g(x, y)\frac{\partial f}{\partial y}+f(x, y)\frac{\partial g}{\partial y}\end{array}$$

Quotient Rule:

$$\begin{array}{l}\text{If}\ u = \frac{f(x,y)}{g(x,y)},\ \text{where}\ g(x, y)\ne 0,\ \text{then}\end{array}$$
$$\begin{array}{l}u_{x}=\frac{g\left ( x,y \right )\frac{\partial f}{\partial x}-f\left ( x,y \right )\frac{\partial g}{\partial x}}{\left [ g\left ( x,y \right ) \right ]^{2}}\end{array}$$

And,

$$\begin{array}{l}u_{y}=\frac{g\left ( x,y \right )\frac{\partial f}{\partial y}-f\left ( x,y \right )\frac{\partial g}{\partial y}}{\left [ g\left ( x,y \right ) \right ]^{2}}\end{array}$$

Power Rule:

If u = [f(x,y)]2 then, partial derivative of u with respect to x and y defined as

$$\begin{array}{l}u_{x} = n\left [ f\left ( x,y \right ) \right ]^{n – 1} \frac{\partial f}{\partial x}\end{array}$$

And,

$$\begin{array}{l}u_{y} = n\left [ f\left ( x,y \right ) \right ]^{n – 1} \frac{\partial f}{\partial y}\end{array}$$

Chain Rule:

If u = f(x,y) is a function, where x = (s,t) and y = (s,t), then by the chain rule, we can find the partial derivatives us and ut as:

$$\begin{array}{l}u_{s}=\frac{\partial u}{\partial x}.\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial s}\end{array}$$

and

$$\begin{array}{l}u_{t}=\frac{\partial u}{\partial x}.\frac{\partial x}{\partial t} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial t}\end{array}$$

## Partial Derivative Examples

Given below are some examples of partial derivatives.

Question 1: Determine the partial derivative of a function fx and fy: if f(x, y) is given by f(x, y) = tan(xy) + sin x.

Solution:

Given function is f(x, y) = tan(xy) + sin x

Derivative of a function with respect to x is given as follows:

$$\begin{array}{l}f_x=\frac{\partial f}{\partial x}\end{array}$$
$$\begin{array}{l}=\frac{\partial}{\partial x}[\tan (xy) + \sin x]\end{array}$$
$$\begin{array}{l}=\frac{\partial}{\partial x}[\tan(xy)] +\frac{\partial}{\partial x}[\sin x]\end{array}$$

= sec2 (xy). y + cos x

= y sec2(xy) + cos x

Now, the derivative of a function with respect to y is as follows:

So, x is the constant,

$$\begin{array}{l}f_y=\frac{\partial f}{\partial y}\end{array}$$
$$\begin{array}{l}=\frac{\partial}{\partial y}[\tan (xy) + \sin x]\end{array}$$
$$\begin{array}{l}=\frac{\partial}{\partial y}[\tan (xy)] +\frac{\partial}{\partial y}[\sin x]\end{array}$$

= sec2 (xy) x + 0

= x sec2 (xy)

Answer: fx = y sec2(xy) + cos x and fy = x sec2 (xy)

Question 2: If f(x,y) = 2x + 3y, where x = t and y = t2, find the derivative f with respect to t.

Solution:

We know,

$$\begin{array}{l}\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}\end{array}$$

Then,

$$\begin{array}{l}\frac{\partial f}{\partial x}=2,\end{array}$$
$$\begin{array}{l}\frac{\partial f}{\partial y}= 3,\end{array}$$
$$\begin{array}{l}\frac{dx}{dt}= 1,\end{array}$$
$$\begin{array}{l}\frac{dy}{dt}= 2t\end{array}$$

So, we get

$$\begin{array}{l}\frac{df}{dt}= 2.1 + 3.(2t) = 2 + 3t\end{array}$$

Question 3: If

$$\begin{array}{l}f = x^{2}(y – z) + y^{2}(z – x) + z^{2}(x – y)\end{array}$$
, prove that
$$\begin{array}{l}\frac {\partial f} {\partial x}+\frac {\partial f} {\partial y}+\frac {\partial f} {\partial z}= 0\end{array}$$
.

Solution:

Given,

$$\begin{array}{l}f = x^{2} (y – z) + y^{2}(z – x) + z^{2}(x – y)\end{array}$$

To find ∂f/∂x, ‘y and z’ are held constant, and the resulting function of ‘x’ is differentiated with respect to ‘x’.

$$\begin{array}{l}\frac {\partial f} {\partial x}=2x(y – z) + 0(0 – 1) + 0(1 – 0)\end{array}$$
$$\begin{array}{l}\frac {\partial f} {\partial x}=2xy – 2xz + 0 + 0\end{array}$$
$$\begin{array}{l}\frac {\partial f} {\partial x}=2xy – 2xz…..(i)\end{array}$$

To find ∂f/∂y, ‘x and z’ is held constant, and the resulting function of ‘y’ is differentiated with respect to ‘y’.

$$\begin{array}{l}\frac {\partial f} {\partial y}=0(1 – z) + 2y(z – x) + 0(0 – 1)\end{array}$$
$$\begin{array}{l}\frac {\partial f} {\partial y}=0 + 2yz – 2yx + 0\end{array}$$
$$\begin{array}{l}\frac {\partial f} {\partial y}=2yz – 2yx…..(ii)\end{array}$$

To find ∂f/∂z, ‘x and y’ are held constant, and the resulting function of ‘z’ is differentiated with respect to ‘z’.

$$\begin{array}{l}\frac {\partial f} {\partial z}=0(0 – 1) + 0(1 – 1) + 2z(x – y)\end{array}$$
$$\begin{array}{l}\frac {\partial f} {\partial z}=0 + 0 + 2zx – 2zy\end{array}$$
$$\begin{array}{l}\frac {\partial f} {\partial z}=2zx – 2zy…..(iii)\end{array}$$

Adding (i), (ii) and (iii), we get

$$\begin{array}{l}\frac {\partial f} {\partial x}+\frac {\partial f} {\partial y}+\frac {\partial f} {\partial z}=2xy – 2xz + 2yz – 2yx – 2zx – 2zy\end{array}$$
$$\begin{array}{l}\frac {\partial f} {\partial x}+\frac {\partial f} {\partial y}+\frac {\partial f} {\partial z}=0\end{array}$$

Hence proved.

Question 4: Given F = sin (xy). Show that ∂2F / (∂x ∂y) is equal to ∂2F / (∂y ∂x).

Solution:

F = sin(xy)

∂F / ∂x = y cos (xy)

∂F / ∂y = x cos (xy)

∂2F / ∂x ∂y = ∂ / ∂y [y cos (xy)]

∂2F / ∂x ∂y = −xy sin (xy) + cos (xy)

∂2F / ∂y ∂x = ∂ / ∂x [x cos (xy)]

∂2F / ∂y ∂x = −xy sin (xy) + cos (xy)

Hence,

∂2F / (∂x ∂y) =∂2F / (∂y ∂x)

Question 5: f (x, y) = x2 + xy + y2 , x = uv, y = u/v. Show that ufu + vfv = 2xfx and ufu − vfv = 2yfy.

Solution: We need to find fu, fv, fx and fy.

fu = ∂f / ∂u = [∂f/ ∂x] [∂x / ∂u] + [∂f / ∂y] [∂y / ∂u];

fv = ∂f / ∂v = [∂f / ∂x] [∂x / ∂v] + [∂f / ∂y] [∂y / ∂v].

fu = (2x + y)(v) + (x + 2y)(1 / v) = 2uv2 + 2u + 2u / v2 .

fv = (2x + y)(u) + (x + 2y)(−u / v2 ) = 2u2 v − 2u2 / v3 .

fx = 2x + y = 2uv + u / v .

So, 2xfx = 2uvfx = 4u2v2 + 2u2 .

fy = x + 2y = uv + 2u / v .

So, 2yfy = [2u / v] fx = 2u2 + 4u2/  v2 .

Now, ufu + vfv = 2u2 v2 + 2u2 + 2u2 / v2 + 2u2 v2 − 2u2 / v2

= 4u2 v2 + 2u2 = 2xfx as required,

and ufu − vfv = 2u2 v2 + 2u2 + 2u2 / v2 − 2u2 v2 + 2u2 / v2

= 2u2 + 4u2 / v2 = 2yfy as required.

Question 6: Show that the largest triangle of the given perimeter is equilateral.

Solution:

A = s (s − a) (s − b) (s − c)

Where,

P = given

s = [1 / 2] P

a+b+c=P

c=P–a–b ← equation (1)

$$\begin{array}{l}A = \sqrt{\frac{1}{2}P(\frac{1}{2}P – a)(\frac{1}{2}P – b)[\frac{1}{2}P – (P – a – b)]}\\ A = \sqrt{\frac{1}{2}P(\frac{1}{2}P – a)(\frac{1}{2}P – b)(a + b – \frac{1}{2}P)}\\ \frac{\partial A}{\partial a} = \frac{\frac{1}{2}P(\frac{1}{2}P – b)\left[ (\frac{1}{2}P – a)(1) + (a + b – \frac{1}{2}P)(-1) \right]}{2\sqrt{\frac{1}{2}P(\frac{1}{2}P – a)(\frac{1}{2}P – b)(a + b – \frac{1}{2}P)}} = 0\\\frac{1}{2}P – a = a + b – \frac{1}{2}P\\\end{array}$$

b = P – 2a ←  equation (2)

$$\begin{array}{l}\frac{\partial A}{\partial b} = \frac{\frac{1}{2}P(\frac{1}{2}P – a)\left[ (\frac{1}{2}P – b)(1) + (a + b – \frac{1}{2}P)(-1) \right]}{2\sqrt{\frac{1}{2}P(\frac{1}{2}P – a)(\frac{1}{2}P – b)(a + b – \frac{1}{2}P)}} = 0\\\frac{1}{2}P – b = a + b – \frac{1}{2}P\\\end{array}$$

a = P – 2b

Substitute b = P – 2a of equation (2)

$$\begin{array}{l}a = P – 2(P – 2a)\\ a = P – 2P + 4a\\ 3a = P\\ a = \frac{1}{3}P\\\end{array}$$

From equation (2)

$$\begin{array}{l}b = P – 2(\frac{1}{3}P)\\ b = \frac{1}{3}P\\\end{array}$$

From equation (1)

$$\begin{array}{l}c = P – \frac{1}{3}P – \frac{1}{3}P\\ c = \frac{1}{3}P\\ \text{Hence,} \ a \ = \ b \ = \ c \ = \ P/3.\end{array}$$

Thus, the triangle is equilateral.

The maximum area of the triangle is:

$$\begin{array}{l}A_{max} = \sqrt{\frac{1}{2}P(\frac{1}{2}P – \frac{1}{3}P)(\frac{1}{2}P – \frac{1}{3}P)(\frac{1}{2}P – \frac{1}{3}P)}\\ A_{max} = \sqrt{\frac{1}{2}P(\frac{1}{6}P)(\frac{1}{6}P)(\frac{1}{6}P)}\\ A_{max} = \sqrt{\frac{1}{432}P^4}\\ A_{max} = \frac{1}{12\sqrt{3}}P^2\\\end{array}$$

Derivative Examples

Differentiation of Functions

## Methods of Differentiation – JEE Solved Questions ## Important Theorems of Differentiation for JEE Q1

### What do you mean by partial differentiation?

Finding the partial derivative of a function is called partial differentiation.

Q2

### When do we use partial differentiation?

We use partial differentiation to differentiate functions having more than one variable.

Q3

### Give the product rule of partial differentiation.

If u = f(x, y).g(x, y), then ∂u/∂x = g(x, y) (∂f/∂x) + f(x, y) (∂g/∂x).
∂u/∂y = g(x, y) (∂f/∂y) + f(x, y) (∂g/∂y).

Q4

### How do we represent the partial derivative of f with respect to y?

The partial derivative of f with respect to y is denoted by ∂f/∂y.