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Students often get anxious about the idea of "calculus". It is a branch of mathematics dealing with the constant change of quantities over a period of time. Differential and Integral Calculus are the two main branches of calculus.
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Calculus is all about understanding how things change over time. It focuses on derivatives and integrals, which essentially involve adding up tiny changes step by step.
Derivatives help us know how one quantity changes in relation to another.
At the same time, integrals enable us to measure the total effect of those changes, such as finding the area under a curve or the distance traveled.
Based on the change and mathematical analysis, calculus can be divided into four parts.
Differential calculus determines the change of a function f(x) with respect to an independent variable. If the function f(x) defines a curve, then its derivatives give the value of its slope.
The derivative of the function is expressed by \({dy \over dx} \text {or f'(x)}\). Here, 'dy' and 'dx' are known as the differentials.
Some important concepts of differential calculus are:
Integrals are the process of calculating the area under the given curve. They are also referred to as the antiderivative of the function. The method of finding the integrals is known as integration.
Integral are further divided into two categories:
\(\int f(x) dx = {F(x)} + c\)
Here,
The integral is the inverse process of the differential. The fundamental theorem of calculus links derivatives and integrals. There are two theorems that are the first and the second fundamental theorems of calculus.
In calculus, for each function, we have different formulas. And students should remember and use the correct formula. For a better understanding of calculus, it's important to remember all the basic formulas.
The derivatives of functions are calculated using different formulas. Children need to memorize these in order to solve complex problems. The formulas are mentioned below:
\( \frac{d}{dx}(x^n) = n x^{n-1} \)
\( \frac{d}{dx}(C) = 0 \quad \text{where $C$ is a constant} \)
\( \frac{d}{dx}(C f) = C \frac{d}{dx}(f) \)
\( \frac{d}{dx}(f \pm g) = \frac{d}{dx}(f) \pm \frac{d}{dx}(g) \)
\( \frac{d}{dx}(f g) = f' g + f g' \)\( \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{g \frac{d}{dx}(f) - f \frac{d}{dx}(g)}{g^2} \)
\( \frac{d}{dx}(\sin x) = \cos x \)
\( \frac{d}{dx}(\cos x) = -\sin x \)
\( \frac{d}{dx}(\tan x) = \sec^2 x \)
\( \frac{d}{dx}(\cot x) = -\csc^2 x \)
\( \frac{d}{dx}(\sec x) = \sec x \tan x \)
\( \frac{d}{dx}(\csc x) = -\csc x \cot x \)
\( \frac{d}{dx}(a^x) = a^x \ln a \)
\( \frac{d}{dx}(e^x) = e^x \)
\( \frac{d}{dx}(\log_a x) = \frac{1}{x \ln a} \)
Integrals are complementary to derivative, hence they are derived from the formula of derivative. The integrals formulas for different functions are as follows:
\( \begin{aligned} &\int 1 \, dx = x + C \\ &\int a \, dx = a x + C \\ &\int \frac{1}{x} \, dx = \ln|x| + C \\ &\int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1} x + C \\ &\int \frac{1}{\sqrt{1 + x^2}} \, dx = \tan^{-1} x + C \\ &\int \frac{1}{|x| \sqrt{x^2 - 1}} \, dx = \sec^{-1} x + C \\ &\int \sin x \, dx = -\cos x + C \\ &\int \cos x \, dx = \sin x + C \\ &\int \sec x \tan x \, dx = \sec x + C \\ &\int \csc x \cot x \, dx = -\csc x + C \\ &\int \sec^2 x \, dx = \tan x + C \\ &\int \csc^2 x \, dx = -\cot x + C \\ &\int e^x \, dx = e^x + C \\ &\int a^x \, dx = \frac{a^x}{\ln a} + C, \quad a>0, a\neq 1 \\ &\int \log x \, dx = x \ln x - x + C \\ &\int \log_a x \, dx = \frac{x \ln x - x}{\ln a} + C \end{aligned} \)
Calculus is a broad topic that includes limits, derivatives, integrals, and many more. Let’s learn some tips and tricks to make it easier for us to learn.
Mistakes are common in calculus, moreover, students repeat the same mistakes. To master calculus, let’s learn some common mistakes.
We have talked a lot about calculus. We know that we use it in our daily life. Now let’s see the applications of calculus.
Find the derivatives of 3x² + 2x + 1.
The derivatives of \(3x^2 + 2x + 1 = 6x + 2\)
\( \frac{d}{dx}(3x^2 + 2x + 1) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x) + \frac{d}{dx}(1) \)
Derivatives of x2 = 2x
Derivatives of 2x = 2
Derivatives of 1 = 0
Add the derivatives: \( \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x) + \frac{d}{dx}(1) \)
\(= 3 × 2x + 2 + 0 \\ \space \\ = 6x + 2\)
Therefore, the derivatives of \(3x^2 + 2x + 1 = 6x + 2\)
Calculate ∫2x dx.
\(∫2x dx = x^2 + c\)
The power rule of integration \(= ∫x^n \ dx = \frac{x^{\ n \ + \ 1}} { n \ + \ 1} + c\)
\(∫2x \ dx = 2 \big [\frac {x^{ \ (1\ + \ 1)} } { 1 \ + \ 1} \big ]+ c\)
\(= 2[ \frac{x^2} { 2}] + c\)
\(= x^2 + c\)
Therefore, \(∫2x dx = x^2 + c\)
Solve ∫(3x² - 4x + 5)dx.
\(∫(3x^2 - 4x + 5)dx = x^3 - 2x^2 + x + c\)
\( \int 3x^2 \, dx = x^3, \quad \int -4x \, dx = -2x^2, \quad \int 5 \, dx = 5x \)
Adding these values:
\( \int 3x^2 \, dx - \int 4x \, dx + \int 5 \, dx \)
\(= \int (3x^2 - 4x + 5) \, dx = x^3 - 2x^2 + 5x + C \)
Where ‘c’ is the constant.
Solve ∫(2x² - 3x - 7)dx.
\( \int (2x^2 + 3x - 7) \, dx = \frac{2x^3}{3} + \frac{3x^2}{2} - 7x + C \)
Solve ∫(4x² - 5x - 6)dx.
\( \int (4x^2 + 5x - 6) \, dx = \frac{4x^3}{3} + \frac{5x^2}{2} - 6x + C \)
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