Differentiation Formulas 

Differentiation is one of the most important topics in calculus, especially for Class 12 students preparing for competitive exams like JEE. It is the process of finding the derivative of a function, which measures the rate of change of a quantity. In this blog, we will cover all differentiation formulas, including basic formulas, advanced rules, and solved examples to help you master differentiation concepts effectively.

1.0What Are Differentiation Formulas?

Differentiation formulas are predefined rules that help us compute the derivative of a function in an easy and systematic way without going back to first principles every time. These formulas are essential for solving problems efficiently in exams.

2.0Basic Differentiation Formulas (Class 12 & JEE)

3.0Important Differentiation Formulas and Rules

1. Product Rule: $\frac{d}{dx}[u(x)\cdot v(x)]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$

2. Quotient Rule: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{v(x)u^{\prime }(x)-u(x)v^{\prime }(x)}{[v(x){]}^2}$

3. Chain Rule: If y = f(g(x)), then: $\frac{dy}{dx}=f^{\prime }(g(x))\cdot g^{\prime }(x)$

4.0Examples of Differentiation Formulas 

Example 1: Differentiate $f(x)=x^5+3x^3-2x+7.$

Solution: $f^{\prime }(x)=5x^4+9x^2-2$

Example 2: Differentiate $f(x)=\sin x\cdot \cos x.$

Solution : $f^{\prime }(x)=(\cos x)(\cos x)+(\sin x)(-\sin x)={\cos }^{2}x-{\sin }^{2}x$

Example 3: Differentiate $f(x)=\sin (3x^2+2x)$

Solution: $f^{\prime }(x)=\cos (3x^2+2x)\cdot (6x+2)=(6x+2)\cos (3x^2+2x)$

Example 4: Differentiate $f(x)=\frac{x^2+1}{x-1}$

Solution:

$f^{\prime }(x)=\frac{(x-1)(2x)-(x^2+1)(1)}{(x-1{)}^2}=\frac{2x(x-1)-(x^2+1)}{(x-1{)}^2}$

Simplify numerator: $x^2-2x-x^2-1=x^2-2x-1$

Thus, $f^{\prime }(x)=\frac{x^2-2x-1}{(x-1{)}^2}$

Example 5: Differentiate $f(x)=\frac{\sin x}{x^2+1}$

Solution:

$f^{\prime }(x)=\frac{(x^2+1)\cos x-\sin x\cdot 2x}{(x^2+1{)}^2}$

So,

$f^{\prime }(x)=\frac{(x^2+1)\cos x-2x\sin x}{(x^2+1{)}^2}$

Example 6: Differentiate $f(x)={\cos }^3(2x+1)$

Solution:

Let $u=\cos (2x+1)$ so $f(x)=u^3$

Step 1 – Apply chain rule:

$f^{\prime }(x)=3u^2\frac{du}{dx}=3{\cos }^2(2x+1)\frac{d}{dx}[\cos (2x+1)].$

Step 2 – Differentiate $\cos (2x+1)$ :

$\frac{d}{dx}[\cos (2x+1)]=-\sin (2x+1)\cdot 2=-2\sin (2x+1).$

Step 3 – Combine:

$f^{\prime }(x)=3{\cos }^2(2x+1)\cdot (-2\sin (2x+1))=-6{\cos }^2(2x+1)\sin (2x+1).$

Example 7: Differentiate $f(x)=\ln (x^2+sinx)$

Solution:

$f^{\prime }(x)=\frac{d}{dx}[\ln (x^2+\sin x)]=\frac{2x+\cos x}{x^2+\sin x}.$

Example 8: Differentiate $f(x)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

Solution:

We know that:$\frac{d}{dx}[ta{n}^{-1}x]=\frac{1}{1+x^2}$

Let $u=\frac{2x}{1-x^2}$

Step 1 – Derivative of u:

$\frac{du}{dx}=\frac{(1-x^2)(2)-(2x)(-2x)}{(1-x^2{)}^2}=\frac{2(1-x^2)+4x^2}{(1-x^2{)}^2}=\frac{2+2x^2}{(1-x^2{)}^2}$

Step 2 – Apply chain rule:

$f^{\prime }(x)=\frac{1}{1+u^2}\frac{du}{dx}$

Simplify $1+u^2$ :

$1+{\left(\frac{2x}{1-x^2}\right)}^2=\frac{(1-x^2{)}^2+(2x{)}^2}{(1-x^2{)}^2}=\frac{1-2x^2+x^4+4x^2}{(1-x^2{)}^2}=\frac{1+2x^2+x^4}{(1-x^2{)}^2}$

Thus, $f^{\prime }(x)=\frac{2+2x^2}{1+2x^2+x^4}$

Example 9: Differentiate $f(x)=e^{x}sinx$

Solution  :

$f^{\prime }(x)=\frac{d}{dx}[e^x\sin x]=e^x\sin x+e^x\cos x$

So,

$f^{\prime }(x)=e^x(\sin x+\cos x)$

Example 10: Find $\frac{d^2}{d{x}^2}[x^3\ln x]$

Solution:

Step 1 – First derivative:

$f^{\prime }(x)=\frac{d}{dx}[x^3\ln x]=3x^2\ln x+x^3\cdot \frac{1}{x}=3x^2\ln x+x^2$

Step 2 – Second derivative:

$f^{\prime \prime }(x)=\frac{d}{dx}[3x^2\ln x+x^2]=3[2x\ln x+x]+2x=6x\ln x+3x+2x=6x\ln x+5x$

So, $f^{\prime \prime }(x)=x(6\ln x+5)$

Example 11: Differentiate implicitly: $x^2+y^2=25$

Solution:

Differentiate both sides w.r.t x: $2x+2y\frac{dy}{dx}=0$

Solving for $\frac{dy}{dx}$

$\frac{dy}{dx}=-\frac{x}{y}$

Example 12: Differentiate $f(x)=(x^2+1{)}^x$

Solution:

Step 1 – Take logarithm on both sides: $\ln f(x)=x\ln (x^2+1)$

Step 2 – Differentiate both sides:

$\frac{f^{\prime }(x)}{f(x)}=\ln (x^2+1)+x\cdot \frac{2x}{x^2+1}=\ln (x^2+1)+\frac{2x^2}{x^2+1}$

Step 3 – Multiply by f(x):

$f^{\prime }(x)=(x^2+1{)}^x\left[\ln (x^2+1)+\frac{2x^2}{x^2+1}\right]$

5.0Practice Questions on Differentiation Formulas

1. Differentiate $f(x)=e^{2x}sinx$
2. Find the derivative of $f(x)=ln(x^2+1)$
3. Differentiate $f(x)=ta{n}^{-1}(3x)$
4. Find $\frac{d}{dx}[x^{3}lnx]$

6.0Sample Question on Differentiation Formulas

Q1. What is the basic differentiation formula for ?

Ans: $\frac{d}{dx}{x}^n=n{x}^{n-1}$