The mathematical study of continuous change we call it calculus. The calculus originally is known as infinitesimal calculus. The study of calculus is the same as geometry is the study of shape. Let us now discuss more calculus and calculus formulas.

**What is Meant by Calculus?**

Calculus is the branch of mathematics, which deals in the study rate of change and its application in solving the equations. Differential calculus and integral calculus are the two major branches of calculus. Differential Calculus deals with the rates of change and slopes of curves.

Integral Calculus deals mainly with the accumulation of quantities and the areas under and between curves. In calculus, we use the fundamental notions of convergence of infinite sequences and infinite series to a well-defined limit. These fundamentals are used by both differential and integral calculus.

In other words, we can say that in differential calculus, an area splits up into small parts to calculate the rate of change. Whereas in integral calculus** **small parts are joined to calculates the area or volume and it is the method of reasoning or calculation.

For calculating very small quantities, we use Calculus. Initially, the first method of doing the calculation with very small quantities was by infinitesimals. For example, an infinitesimal number could be greater than 0, but less than any number in the sequence 1, 1/2, 1/3, etc.

In other words, its value is less than any positive real number. According to this point of view, we can say that calculus is a collection of techniques for manipulating infinitesimals. In calculus the symbols \({\displaystyle dx}dx and {\displaystyle dy}dy\) were taken to be infinitesimal, and the derivative \(\frac{dy}{dx}{\displaystyle dy/dx}\) was simply their ratio.

**Differential Calculus**

The study of the definition, properties, and applications of the derivative of a function is known as Differential calculus. Differentiation is the process of finding the derivative. Here are some calculus formulas by which we can find derivative of a function.

- \(\frac{\mathrm{d} r^2}{\mathrm{d} x} = nx^(n−1)\)
- \(\frac{\mathrm{d} (fg) }{\mathrm{d} x} = fg^1+gf^1\)
- \(\frac{\mathrm{d} }{\mathrm{d} x} (\frac{f}{g})\)= \(\frac{ gf^1−fg^1}{ g^2}\)
- \(\frac{\mathrm{d} f(g(x))}{\mathrm{d} x} = f^1(g(x))g^1(x)\\\)
- \(\frac{\mathrm{d} (sinx) }{\mathrm{d} x} =cosx\)
- \(\frac{\mathrm{d} (cosx)}{\mathrm{d} x} =−sinx\)
- \(\frac{\mathrm{d} (tanx)}{\mathrm{d} x} =−sec2x\)
- \(\frac{\mathrm{d} (cotx)}{\mathrm{d} x} =csc2x\)
- \(\frac{\mathrm{d} (secx)}{\mathrm{d} x} =secxtanx\)
- \(\frac{\mathrm{d} (cscx) }{\mathrm{d} x} =−cscxcotx\)
- \(\frac{\mathrm{d} (e^x) }{\mathrm{d} x} =e^x\)
- \(\frac{\mathrm{d} (a^x)}{\mathrm{d} x} =a^x \ln a\)
- \(\frac{\mathrm{d} }{\mathrm{d} x}\ln x= \frac{1}{x}\)
- \(\frac{\mathrm{d}(arcsinx) }{\mathrm{d} x} = \frac{1}{\sqrt{1 – x^2}}\)
- \(\frac{\mathrm{d} (arcsinx)}{\mathrm{d} x} = \frac{1}{\sqrt{1 + x^2}}\)

**Integration Formulas**

The branch of calculus where we study about integrals, accumulation of quantities and the areas under and between curves and their properties is known as Integral Calculus. Here are some formulas by which we can find integral of a function.

- \(\int a dr=ax+C\)
- \(\int \frac{1}{x} dr=\ln \left | x \right |+C\)
- \(\int ax dx= \frac{ex}{\ln a} +C\)
- \(\int \ln x dx=x \ln x−x+C\)
- \(\int sinx dx=−cosx+C\)
- \(\int cosx dx=sinx+C\)
- \(\int tanx dr+\ln \left | secx \right |+C = −\ln \left | cosx \right | +C\)
- \(\int cotx dr=\ln \left | sin x \right |+C\)
- \(\int secxdx=\ln \left | secx+tanx \right |+C\)
- \(\int cscxdx=\ln \left |cscx–cotx \right |+C\)
- \(\int sec^2xdx=tanx+C\)
- \(\int secxtanxdx=secx+C\)
- \(\int csc^2xdr=−cotx+C\)
- \(\int tan^2xdr=tanx−x+C\)
- \(\int frac{dr}{\sqrt{a^2−x^2}} =arcsin\frac{x}{a} +C\)
- \(\int frac{dr}{\sqrt{a^2+x^2}} = \frac{1}{a}arcsin\frac{x}{a}+C\)

## Solved Examples for Calculus Formulas

Q 1: Find \(\int cos a + a d(a)\)

Solution: \(\int cos a + a d(a) = \int cos a da + \int a d(a)\)

= \(sin a + \frac{x^2}{2}+ c\)