Calculus! Many students get fear with this branch of mathematics, as it
is considered as one of the difficult topics of math. The word Calculus
is derived from Latin word, which means calculating something and used
for counting anything. Calculus covers a wide area of modern mathematics. It mainly focuses on limits, functions, derivatives, integrals, and
infinite series. Calculus plays a very important role in modern
mathematics. It has two branches Differential Calculus and Integral
Calculus. |

__Chain rule is a formula for computing the derivative of the composition of two or more functions.__

**Chain rule:**If l and m are two functions, then the chain rule is

$\frac{dl}{dy}$ = $\frac{dl}{dm}$ . $\frac{dm}{dy}$

__where, 'l' differentiate with respect to m and m differentiate with respect to y.__

__A function with no gaps, jumps or undefined points. A continuous function is a function in which small changes in the input results to small changes in the output.__

**Continuity:**__Let f be a function which is continuous on the closed interval [a, b]. The definite integral of f from a to b is defined to be the limit__

**Definite Integral:**$\int_{a}^{b}$ f(x) dx = $\lim_{n\rightarrow \infty}$ $\sum_{i=1}^{n}$f($x_{i})$ $\Delta x$

**Derivative is a measure of how a function changes as its input changes.**

__Derivative:____If a function is not continuous at a point in its domain, then it is said to have a discontinuity.__

**Discontinuity:**__Helps in determining the possible maximum and minimum values of a function on certain intervals.__

**Extreme Value Theorem:**If a function f(x) is continuous on a closed interval [a, b], then f(x) has both a maximum and minimum values on [a, b].

**Any derivative beyond the first derivative is referred as higher order derivative.**

__Higher Order Derivatives:__**An integral without upper and lower limits.**

__Indefinite Integral:__**The value of a function approaches as the variable approaches some point. If the function is not continuous, the limit could be different from the value of the function at that point.**

__Limit:____Uses derivatives to evaluate limits involving indeterminate forms.__

**L Hospital's Rule:**$\lim_{x\rightarrow c}$ $\frac{f(x)}{g(x)}$ = $\lim_{x\rightarrow c}$$\frac{f'(x)}{g'(x)}$

__A function f(x) has relative maximum value at x = a, if f(a) is greater than any value immediately preceding or following.__

**Maximum:****If f(x) is defined and continuous on the interval [a, b] and differentiable on (a, b), then there is atleast one number c in the interval (a, b), such that**

__Mean Value Theorem:__f'(c) = $\frac{f(b) - f(a)}{b-a}$

**A function f(x) has relative minimum value at x = b, if f(b) is less than any value immediately preceding or following.**

__Minimum:__**Method for finding successively better approximations to the roots of a real-valued function.**

__Newton's Method:__**Slope is the ratio of the rise divided by the run between two points on a line. It describes its steepness, incline or grade.**

__Slope:__**Power rule is an important differentiation rule. Polynomials are differentiated using the rule.**

__Power Rule:__$\frac{d}{dx}$ x$^{n}$ = nx$^{n - 1}$ ; n$\neq$ 0

__Method of finding the derivative of a function that is the quotient of two other functions for which derivative exist.__

**Quotient Rule:**Quotient rule is given as

$\frac{d}{dx}$ ($\frac{g(x)}{h(x)})$ = $\frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^{2}}$ Given below are some of the problems in calculus.

### Solved Examples

**Question 1:**Is the following function continuous?

f(x) = $\begin{Bmatrix}

x^{2}-1&for\ x<2 \\

x^{3}+5 & for\ x\geq2

\end{Bmatrix}$

**Solution:**

$\lim_{x\rightarrow 2^{-}}$f(x)= $\lim_{x\rightarrow 2^{-}}(x^{2}-1)$ = 4 - 1 = 3

$\lim_{x\rightarrow 2^{+}}$f(x)= $\lim_{x\rightarrow 2^{+}}(x^{3}+5)$ = 8 + 5 = 13

Since $\lim_{x\rightarrow 2^{-}}$ f(x) $\neq$ $\lim_{x\rightarrow 2^{+}}$

f(x) does not exist and f is not continuous at x = 2.

**Question 2:**Solve $\lim_{x\rightarrow 3}$ $\frac{3x^{2}+3x -5}{3-2x}$

**Solution:**

= $\frac{\lim_{x\rightarrow 3}3x^{2}+\lim_{x\rightarrow 3}3x - \lim_{x\rightarrow 3} 5}{\lim_{x\rightarrow 3}3 -\lim_{x\rightarrow 3}2x }$

= $\frac{3(3)^{2}+ 3(3)-5}{3 - 2(3)}$

= $\frac{27+9-5}{3-6}$

= - $\frac{31}{3}$