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The process of finding the derivative or the differential coefficient of a function w.r.t. the variable, on which it depends is called differentiation. Integration is the process of finding the function, whose derivative is given. for this reason, the process of integration is called the inverse process of differentiation.

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This Story also Contains

1. Differentiation
2. Integration
3. Solved Example Based on Differentiation and Integration
4. Conclusion

In this article, we will cover the mathematical tools used in kinematics. These tools are crucial for studying any chapter, especially in the mechanics section, where a solid understanding of basic mathematics is essential for solving problems.

So let’s read the entire article to know in-depth about the differentiation and integration to use in kinematics to make the calculation easier.

Differentiation

Differentiation is very useful when we have to find rates of change of one quantity compared to another.

• If y is one quantity and we have to find the rate of change of y with respect to x which is another quantity

$\text{Then the differentiation of}\mathrm{y}\text{w.r.t}\mathrm{x}\text{is given as}\frac{dy}{dx}$

• For a y V/s x graph

We can find the slope of the graph using differentiation

I.e Slope of $\mathrm{y}\mathrm{V}/\mathrm{s}\times$ graph $=\frac{dy}{dx}$
- Some important Formulas of differentiation
- $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Example-

$\begin{array}{rl}& \frac{d}{dx}\left(x^5\right)=(n=5)\\ & \because \frac{\mathrm{d}{x}^n}{\mathrm{d}x}=n{x}^{n-1}\\ & \therefore \frac{\mathrm{d}{x}^5}{\mathrm{d}x}=5x^{5-1}\\ & \Rightarrow \frac{\mathrm{d}{x}^5}{\mathrm{d}x}=5x^4\end{array}$

Similarly

$\begin{array}{rl}& \frac{d}{dx}\sin x=\cos x\\ & \frac{d}{dx}\cos x=-\sin x\\ & \frac{d}{dx}\tan x={\sec }^{2}x\\ & \frac{d}{dx}\cot x=-{\csc }^{2}x\\ & \frac{d}{dx}\sec x=\sec x\tan x\\ & \frac{d}{dx}\csc x=-\csc x\cot x\\ & \frac{d}{dx}{e}^x=e^x\\ & \frac{d}{dx}{a}^x=a^x\ln a\\ & \frac{d}{dx}\ln |x|=\frac{1}{x}\end{array}$

Integration

The opposite process of differentiation is known as integration.
Let x , and y be two quantities, using differentiation we can find the rate of change of y with respect to x , Which is given by $\frac{dy}{dx}$

But using integration we can get the direct relationship between quantities x and y

So let $\frac{dy}{dx}=K$ where $\mathrm{K}$ is constant
Or we can write $dy=Kdx$

Now integrating on both sides we get the direct relationship between x and y

$\begin{array}{rl}& \text{І.e}\int dy=\int Kdx\\ & y=Kx+C\end{array}$

Where $\mathrm{C}$ is some constant

• For a y V/s x graph

We can find the area of the graph using the integration

Some Important Formulas of Integration

$\begin{array}{rl}& \text{.}\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\text{where (C = constant)}\\ & \text{E.g-}\int x^{n}dx=,n=3\\ & \Rightarrow \frac{x^{n+1}}{n+1}+C\\ & \Rightarrow \frac{x^{3+1}}{3+1}+C\\ & \Rightarrow \frac{x^4}{4}+C\\ & \int \frac{dx}{x}=\ln |x|+C\\ & \int e^{x}dx=e^x+C\\ & \int a^{x}dx=\frac{1}{\ln a}{a}^x+C\\ & \text{.}\int \ln xdx=x\ln x-x+C\end{array}$

$\begin{array}{rl}& \int \sin xdx=-\cos x+C\\ & \int \cos xdx=\sin x+C\\ & \int \tan xdx=-\ln |\cos x|+C\\ & \int \cot xdx=\ln |\sin x|+C\\ & \int \sec xdx=\ln |\sec x+\tan x|+C\\ & \int \csc xdx=-\ln |\csc x+\cot x|+C\\ & \int {\sec }^{2}xdx=\tan x+C\\ & \int {\csc }^{2}xdx=-\cot x+C\\ & \int \sec x\tan xdx=\sec x+C\\ & \int \csc x\cot xdx=-\csc x+C\\ & \int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C\\ & \int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C\\ & \int \frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}{\sec }^{-1}\frac{|x|}{a}+C\end{array}$

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Solved Example Based on Differentiation and Integration

Example 1. Displacement of a particle as a function of time is given as $s=a{t}^3+b{t}^2+c$. Then rate of Displacement as a function of time is
1) Independent of $t$
2) 3at
3) $2bt$
4) $3a{t}^2+2bt$

Solution:

The displacement of a particle as a function of time is given as
$s=a{t}^3+b{t}^2+c\text{.}$

The rate of Displacement-
$\frac{ds}{dt}=\frac{d}{dt}[a{t}^3+b{t}^2+c]=3a{t}^2+2bt$

Hence, the answer is option (1).

Example 2. If $y=x^3$ then $\frac{dy}{dx}$ is:
1) $3x^2$
2) $3x$
3) 3
4) $3x^3$

Solution:

The formula of differentiation
$\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Given - $y=x^3$
So using the formula we get
$\frac{dy}{dx}=3x^2$

Hence, the answer is option (1).

Example 3. Then what is the relation between $x$ and $y$ then what is the relation between $x$ and $y$
1) $y=\frac{x^3}{3}$
2) $y=\frac{x^2}{2}$
3) $y=\frac{x^4}{4}$
4) $y=x$

Solution:

The formula of integration-

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\text{where}(\mathrm{C}=\text{constant})$

Given-
$\begin{array}{rl}& \frac{dy}{dx}=x^2\\ \Rightarrow & dy=x^{2}dx\\ \Rightarrow & y=\int x^{2}dx\\ \Rightarrow & y=\frac{x^3}{3}+c\end{array}$

As at $x=0,y=0$
so, $y=\frac{x^3}{3}$
Hence, the answer is option (1).

Example 4. The distance travelled by an accelerated particle of mass $M$ is given by the following relation $s=6t+3t^2$. Then the rate of change of $s$ after 2 seconds is
1) 18
2) 12
3) 6
4) 24

Solution:

As we learned
Some important Formulas of differentiation
$\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

So $s=6t+3t^2$
now using the formula
$\begin{array}{rl}& ds/dt=6+6t\\ & \text{put}t=(2\sec )\\ & \Rightarrow ds/dt=6+(6\times 2)=18\end{array}$

Hence, the answer is option (1).

Example 5. If $\sin \theta =1/3$, then find the value of $\cos \theta$
1) $\frac{8}{9}$
2) $\frac{4}{3}$
3) $\frac{2\sqrt{2}}{3}$
4) $\frac{3}{4}$

Solution:

$\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-(1/3{)}^2}=\sqrt{8/9}=\frac{2\sqrt{2}}{3}$

Hence, the answer is the option (3).

Conclusion

The derivative or differential coefficient of a function is the limit to which the ratio of the small increment in the function to the corresponding small increment in the variable (on which it depends) tends to, when the small increment in the variable approaches zero. The concept of integration is used in physics to make measurements when a physical quantity varies in continuous manner.