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# Differential Calculus

Top
 Sub Topics Calculus is divided into two branches. One is differential calculus, another is integral calculus. Differential calculus deals with derivatives of functions and their applications. Derivative of a functions is defined as the rate of change of a function. Derivative of a function tells how the value of a function changes with any change in its input. The process of finding derivative is called differentiation. Differential calculus is widely used in physics also. For example: Rate of change of displacement i.e. derivative of displacement of an object is called its velocity and the rate of change of velocity i.e. derivative of the velocity of an object is called its acceleration.

## Differentiation Definition

Differentiation is the method of finding derivative. Derivative is defined as the rate of change of dependent variable y with respect to the change in independent variable x. Let us consider a function f. Then, derivative of function f with respect to independent variable x is denoted by $\frac{\mathrm{d} f}{\mathrm{d} x}$ or by ${f}'(x)$. If we assume a small change (say h or $\Delta x$) in position of x, then according to the definition of differentiation:

## Differentiation Formulas

Few important formulas of differentiation are listed below:
$\frac{\mathrm{d}c }{\mathrm{d} x}$ = 0, where c a constant

$\frac{\mathrm{d} }{\mathrm{d} x}$$x^{n} = nx^{n - 1} \frac{\mathrm{d} }{\mathrm{d} x}$$e^{x} = e^{x}$

$\frac{\mathrm{d} }{\mathrm{d} x}$$a^{x} = a^{x} \ln a, for a > 0 \frac{\mathrm{d} }{\mathrm{d} x}$$\ln x$ = $\frac{1}{x}$

$\frac{\mathrm{d} }{\mathrm{d} x}$$\sin x = \cos x \frac{\mathrm{d} }{\mathrm{d} x}$$\cos x = -\sin x$

$\frac{\mathrm{d} }{\mathrm{d} x}$$\tan x = \sec ^{2}x \frac{\mathrm{d} }{\mathrm{d} x}$$\cot x = -\csc ^{2}x$

$\frac{\mathrm{d} }{\mathrm{d} x}$$\sec x = \sec x \tan x \frac{\mathrm{d} }{\mathrm{d} x}$$\csc x = -\csc x \cot x$

$\frac{\mathrm{d} }{\mathrm{d} x}$$\sin^{-1}x = \frac{1}{\sqrt{1 - x^{2}}} \frac{\mathrm{d} }{\mathrm{d} x}$$\cos^{-1}x = -$$\frac{1}{\sqrt{1 - x^{2}}} \frac{\mathrm{d} }{\mathrm{d} x}$$\tan^{-1}x$ = $\frac{1}{1 + x^{2}}$

$\frac{\mathrm{d} }{\mathrm{d} x}$
$\cot ^{-1}x = -$$\frac{1}{1 + x^{2}} \frac{\mathrm{d} }{\mathrm{d} x}$$\sec ^{-1}x$ = $\frac{1}{x\sqrt{x^{2} - 1}}$

$\frac{\mathrm{d} }{\mathrm{d} x}$$\csc ^{-1}x = -$$\frac{1}{x\sqrt{x^{2} - 1}}$

## Applications of Differentiation

Differentiation is a widely used concept. It is not limited to finding derivatives and solving problems. Along with mathematics, differentiation is applied to physics, chemistry, economics and many other fields. Most commonly used applications of differentiation are described below:
• Differentiation is used to find extrema, which includes maximum and minimum values of a function. For example: Finding the maximum volume of a sphere that can fit into a square of given side.
• Differentiation is used in finding the limit of a function. Sometimes, when we get indeterminate form, L'hospital rule is applied which is based on differentiation.
• In physics, various phenomena are based upon differentiation. For example: Finding velocity, acceleration etc. with which most of the physics concepts are related.
• Differentiation is widely used in economics in determining marginal cost, marginal profit etc.
Application of differentiation has a vast area. There are many more applications of differentiation in commerce, finance, civil engineering, construction of mega structure etc.

## Maxima and Minima

A function can have peaks and valleys or ups and downs. A value at which a given function reaches its relative maximum point, is called its local maximum. Similarly, a value at which a given function reaches its relative minimum point, is called its local minimum. On the other hand, global maximum is a point, where the function reaches its highest point and global minimum is a point where the function reaches its lowest point.

A function can have more than one maximum (plural - maxima) or minimum (plural - minima) as shown in the figure:

### How to Find Maxima or Minima?

Following steps should be followed while finding local maximum and local minimum for a function ${f}(x)$:
• Find first derivative of the function ${f}'(x)$.
• Equate it to zero and determine critical point (say "a"). Critical point is a point, where either maximum or minimum exists.
• Calculate second derivative ${f}''(x)$.
• Evaluate ${f}''(a)$
• If ${f}''(a) < 0$, then there is a maximum at point a.
• If ${f}''(a) > 0$, then there is a minimum at point a.
Let us consider an example.

### Solved Example

Question: Find the maximum or minimum point for a function $f(x) = 3x^{2} - 2x - 1$.
Solution:
First, let us find the first derivative.
${f}'(x) = 6x - 2$

Now, let us find the critical point.
${f}'(x) = 0$
$6x - 2 = 0$
$x$ = $\frac{1}{3}$

Let us find the second derivative.
${f}''(x) = 6 > 0$
Therefore, there is a local minimum at point $x$ = $\frac{1}{3}$.

## Differentiation Examples

Question 1: Find the derivative of the function $y = x^{4} - 3x^{3} - 2\sqrt{x} + $$\frac{2}{7} with respect to x. Solution: y = x^{4} - 3x^{3} - 2\sqrt{x} +$$\frac{2}{7}$
$y = x^{4} - 3x^{3} - 2x$$^{\frac{1}{2}} + \frac{2}{7} \frac{\mathrm{d} y}{\mathrm{d} x} = 4x^{3} - 3 \times 3 \times x^{2} - 2 \times$$\frac{1}{2}x^{-\frac{1}{2}}$$+ 0 \frac{\mathrm{d} y}{\mathrm{d} x} = 4x^{3} - 9x^{2} -$$\frac{1}{\sqrt{x}}$
Question 2: Find $\frac{\mathrm{d} f}{\mathrm{d} x}$, if $f(x) = \sec x - \tan x$.
$f(x) = \sec x - \tan x$
$\frac{\mathrm{d} f}{\mathrm{d} x}$ = $\tan x \sec x - \sec ^{2}x$
$= \sec x(\tan x - \sec x)$