MATHEMATICS: Introduction to calculus formula sheet

Calculus is a branch in mathematics which deals with the study of limits, functions, derivatives, integrals, and infinite series. It has two main branches

Differential calculus
Integral calculus

It is a formula sheet which contains differentiation and integration formulas which are needed to solve the calculus problems. We should know all the calculus formulas before get into the problems. Calculus formula sheet helps you to learn all those formulas. It is very much helpful for the students at the time of examination for reviewing calculus formulas.

Integration is a limiting process which is used to find the area of a region under a curve. We can also say that integration is an anti derivative of differentiation. Integration of a function is shown as,

$\displaystyle\int$ f(x) dx = F(x) + C

$\displaystyle\int$ = Sign of integration

The variable x in dx is called variable of integration or integrator.

C= constant

Differential Calculus Formula Sheet:

Derivatives of polynomial functions:

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (c) = 0 C - Constant

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (x) = 1

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cx) = c

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (xⁿ) = nx^n-1

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cxⁿ) = ncx^n-1

Derivatives of trigonometric functions:

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (sin x) = cos x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cos x) = - sin x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (tan x) = sec²x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cot x) = - cosec²x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (sec x) = sec x tan x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cosec x) = - cosec x cot x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ e^ax = ae^x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ c f(x)=c d/dx f(x)

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ sinax = acosax

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ cosax = −asinax

Basic Rules Necessary to Solve Differential Calculus Problems

Sum Rule for Solve Differential Calculus Problems:

             [f(x)±g(x)]′=f′(x)±g′(x)
Product Rule for Solve Differential Calculus Problems:

             [f(x)·g(x)]′=f′(x)·g(x)+f(x)·g′(x).
Quotient Rule for Solve Differential Calculus Problems:

             [f(x)g(x)]′ = f′(x)·g(x)−f(x)·g′(x) / g2(x)
Chain Rule for Solve Differential Calculus Problems:

             [f•g(x)]′=f′(g(x))·g′(x)
Inverse Rule for Solve Differential Calculus Problems:

      [f⁻¹]′(t)=1 / f′(f⁻¹(t))

ie,

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (u + v) = $\displaystyle\frac{{{d}{u}}}{{\left.{d}{x}\right.}}$ + $\displaystyle\frac{{{d}{v}}}{{\left.{d}{x}\right.}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (u - v) = $\displaystyle\frac{{{d}{u}}}{{\left.{d}{x}\right.}}$ - $\displaystyle\frac{{{d}{v}}}{{\left.{d}{x}\right.}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (uv) = u $\displaystyle\frac{{{d}{v}}}{{\left.{d}{x}\right.}}$ + v $\displaystyle\frac{{{d}{u}}}{{\left.{d}{x}\right.}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (u/v) = (v $\displaystyle\frac{{{d}{u}}}{{\left.{d}{x}\right.}}$ - u $\displaystyle\frac{{{d}{v}}}{{\left.{d}{x}\right.}}$ )/v²

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}$ (e^v) = e^v $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}$ (a^v) = a^v(ln a) $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}$ (ln u) = $\displaystyle\frac{{1}}{{u}}$ $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}$ (log_x u) = $\displaystyle\frac{{1}}{{{\left({\ln{{a}}}\right)}{u}}}$ $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$

Derivatives of inverse trigonometric functions:

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (sin^-1 x) = $\displaystyle\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cos^-1 x) = - $\displaystyle\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (tan^-1 x) = $\displaystyle\frac{{1}}{{{1}+{{x}}^{{2}}}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (sec^-1 x) = $\displaystyle\frac{{1}}{{{\left|{x}\right|}\sqrt{{{{x}}^{{2}}-{1}}}}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cosec^-1 x) = - $\displaystyle\frac{{1}}{{{\left|{x}\right|}\sqrt{{{{x}}^{{2}}-{1}}}}}$

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cot^-1 x) = - $\displaystyle\frac{{1}}{{{1}+{{x}}^{{2}}}}$

Derivatives of hyperbolic functions:

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (sinh x) = cos hx

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cosh x) = sin hx

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (tanh x) = sec h²x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (sech x) = - tanh x sech x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cosech x) = - coth x cosech x

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (coth x) = - cosech²x

Learn Differentiation Revision Problems:

Learn differentiation revision problem 1:

Through the revision of differentiation formula, Find the given function y = 5cot x - x³ .

Solution:

Let y = 5cot x - x³ .

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}\right.}}$ = $\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ [5cot x - x³]

= 5 $\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cot x) - 3x² $\displaystyle\frac{{{\left.{d}{x}\right.}}}{{\left.{d}{x}\right.}}$ (we know . $\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (cot x) = - cosec²x)

= 5(-cosec²x) - 3x²

= - 5cosec²x -3x²

Answer: Differentiation of 5cot x - x³ is - 5cosec²x - 3x²

Learn differentiation revision problem 2:

Through the revision of differentiation formula, Find the exponential function 9e^-4x

Solution:

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (9e^-4x) = 9 $\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ e^-4x

Here, a = -4

So, = 9 $\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ e^-4x

= 9(-4) e^-4x

= -36 e^-4x

Answer: Differentiation of 9e^-4x is -36 e^-4x

Learn differentiation revision problem 3:

Through the revision of differentiation formula, Find the logarithmic function f(x) = ln(4x) with respect to x.

Solution:

Let u = 4x So, f(x) = ln u

$\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$ = 4

f(x) = ln u

$\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (f(x)) = $\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ ( ln u) we know, $\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}$ (ln u) = $\displaystyle\frac{{1}}{{u}}$ $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$

= $\displaystyle\frac{{1}}{{u}}$ $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$

= $\displaystyle\frac{{1}}{{{4}{x}}}{\left({4}\right)}$

= $\displaystyle\frac{{1}}{{{x}}}$

Answer: Differentiation of ln(4x) is $\displaystyle\frac{{1}}{{{x}}}$

Integral Calculus Formula Sheet:

List of integrals of rational functions:

$\displaystyle\int$ k dx = kx + C

$\displaystyle\int$ x^a dx = $\displaystyle\frac{{{{x}}^{{{a}+{1}}}}}{{{a}+{1}}}$ + C

$\displaystyle\int$ 1/x dx = ln|x| + C

List of integrals of logarithmic functions:

$\displaystyle\int$ ln x dx = x ln x - x + C

$\displaystyle\int$ log_ax dx = xlog_ax - x(ln a) + C

List of integrals of exponential functions:

$\displaystyle\int$ e^x dx = e^x + C

$\displaystyle\int$ a^x dx = $\displaystyle\frac{{{{a}}^{{x}}}}{{{\ln{{a}}}}}$ + C

List of integrals of trigonometric functions:

$\displaystyle\int$ sin x dx = - cos x + C

$\displaystyle\int$ cos x dx = sin x + C

$\displaystyle\int$ tan x dx = - ln |cos x| + C

$\displaystyle\int$ cot x dx = ln |sin x| + C

$\displaystyle\int$ sec x dx = ln |sec x + tan x| + C

$\displaystyle\int$ csc x dx = ln |csc x - cot x| + C

List of integrals of inverse trigonometric functions:

$\displaystyle\int$ sin^-1 x dx = x sin^-1 x + $\displaystyle\sqrt{{{1}-{{x}}^{{2}}}}$ + C

$\displaystyle\int$ cos^-1 x dx = x cos^-1 x - $\displaystyle\sqrt{{{1}-{{x}}^{{2}}}}$ + C

$\displaystyle\int$ tan^-1 x dx = x tan^-1 x - $\displaystyle\frac{{1}}{{2}}$ ln |1 + x²| + C

$\displaystyle\int$ cot^-1x dx = x cot^-1 x + $\displaystyle\frac{{1}}{{2}}$ ln |1 + x²| + C

$\displaystyle\int$ sec^-1 x dx = x sec^-1x- cosh^-1 x + C

$\displaystyle\int$ csc^-1 x dx = x csc^-1 x + cosh^-1 + C

List of integrals of hyperbolic functions:

$\displaystyle\int$ sinh x dx = cosh x + C

$\displaystyle\int$ cosh x dx = sinh x + C

$\displaystyle\int$ tanh x dx = ln |coshx| + C

$\displaystyle\int$ csch x dx = ln |tanh $\displaystyle\frac{{x}}{{2}}$ | + C

$\displaystyle\int$ sech x dx = sin^-1 (tanh x) + C

$\displaystyle\int$ coth x dx = ln |sinh x| + C

Other Integration Formula:

$\displaystyle\int\frac{{\left.{d}{x}\right.}}{{\sqrt{{{{x}}^{{2}}+{1}}}}}$ = arcsinh x + C

$\displaystyle\int\frac{{\left.{d}{x}\right.}}{{\sqrt{{{{x}}^{{2}}-{1}}}}}$ = arccosh x + C

$\displaystyle\int\frac{{\left.{d}{x}\right.}}{{{1}-{{x}}^{{2}}}}$ = arctanh x + C, |x| < 1

$\displaystyle\int\frac{{\left.{d}{x}\right.}}{{{1}+{{x}}^{{2}}}}$ = arccoth x + C, |x| < 1

$\displaystyle\int\frac{{\left.{d}{x}\right.}}{{{x}\sqrt{{{1}-{{x}}^{{2}}}}}}$ = -arcsech x + C

$\displaystyle\int\frac{{\left.{d}{x}\right.}}{{{x}\sqrt{{{1}+{{x}}^{{2}}}}}}$ = -arccsch x + C

$\displaystyle\int\frac{{\left.{d}{x}\right.}}{{\sqrt{{{{x}}^{{2}}\pm{{a}}^{{2}}}}}}$ = ln [x + (x² $\displaystyle\pm$ a²)1/2] + C

$\displaystyle\int\frac{{\left.{d}{x}\right.}}{{{{a}}^{{2}}-{{x}}^{{2}}}}=\frac{{a}}{{2}}{\ln{{\left(\frac{{{a}+{x}}}{{{a}-{x}}}\right)}}}+{c}$

$\displaystyle\int\frac{{\left.{d}{x}\right.}}{{{x}\sqrt{{{{a}}^{{2}}\pm{{x}}^{{2}}}}}}=\frac{{a}}{{2}}{\ln{{\left(\frac{{{a}+\sqrt{{{{a}}^{{2}}\pm{{x}}^{{2}}}}}}{{{l}{x}{l}}}\right)}}}+{c}$

$\displaystyle\int\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\left[{{a}}^{{2}}{\arcsin{{\left(\frac{{x}}{{a}}\right)}}}+{x}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right]}+{c}$

$\displaystyle\int\sqrt{{{{x}}^{{2}}\pm{{a}}^{{2}}}}=\frac{{1}}{{2}}{\left[{x}\sqrt{{{{x}}^{{2}}\pm{{a}}^{{2}}}}\pm{{a}}^{{2}}{\ln{{\left({x}+\sqrt{{{{x}}^{{2}}\pm{{a}}^{{2}}}}\right)}}}\right]}+{c}$

Important Integrals Calculus Formulas:

$\displaystyle\int$ ln x dx = x ln x - x + c
$\displaystyle\int$ a^x dx = $\displaystyle\frac{{{{a}}^{{x}}}}{{{\ln{{a}}}}}$ + c
$\displaystyle\int$ xⁿ = $\displaystyle\frac{{{{x}}^{{n}}+{1}}}{{{n}+{1}}}$ + c
$\displaystyle\int\frac{{1}}{{x}}$ dx = ln |x| + c
$\displaystyle\int$ a dx = x + c
$\displaystyle\int$ e^x dx = e^x + c
$\displaystyle\int$ sin x dx = -cosx + c
$\displaystyle\int$ cos x dx = sin x + c
$\displaystyle\int$ cot x dx = ln |sin x| + c
$\displaystyle\int$ sec x dx = ln |sec x + tan x| + c
$\displaystyle\int$ csc x dx = ln|csc x - cot x| + c
$\displaystyle\int$ sec² x dx = tan x + c
$\displaystyle\int$ tan² x dx = tan x - x + c
$\displaystyle\int$ sec x tan x dx = sec x + c
$\displaystyle\int$ csc² x dx = - cot x + c

Important Differential Calculus Formulas:

D' {xⁿ} = nx^n-1
D'{cos x} = -sin x
D'{sin x} = cos x
D'{tan x} = sec² x
D'{sec x} = sec x * tan x
D'{cot x} = - cosec² x
D'{log_ax} = $\displaystyle\frac{{1}}{{x}}$ log_ax
D'{cosec x} = -cosec x * cot x
D'{log x} = $\displaystyle\frac{{1}}{{x}}$
D'{a^x} = a^x log a
D'{e^x} = e^x
D'{sin-¹ x} = $\displaystyle\frac{{{1}}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}$
D'{cos^-1 x} = $\displaystyle\frac{{-{1}}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}$
D'{tan^-1x} = $\displaystyle\frac{{{1}}}{{{1}+{{x}}^{{2}}}}$
D'{cot^-1 x} = $\displaystyle\frac{{-{1}}}{{{1}+{{x}}^{{2}}}}$

List of Calculus Formulas-some Common Derivatives and Integrals.

We know the formulae for the derivatives of many important functions.From these formulae, we can write down immediately the corresponding formulae(referred to as standard formulae) for the integrals of these functions, as listed below which will be used to find integrals of other functions.