Calculus! Many students fear this branch of mathematics, as it is considered as one of the difficult topics of math. The word Calculus is derived from a 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.__

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

**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

$\frac{d}{dx}$ ($\frac{g(x)}{h(x)})$ = $\frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^{2}}$

### 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:**

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

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

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

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