Differentiation Rules: Definition, Formula, Examples

Differentiation is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of differentiation have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

In this article, we will cover the concept of the Rule of Differentiation. This concept falls under the broader category of Calculus, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last five years of the JEE Main exam (from 2013 to 2023), a total of nine questions have been asked on this concept, including one in 2013, one in 2018, two in 2019, three in 2020, one in 2021, and one in 2022.

Rules of Differentiation

The important rules of differentiation are:

• Power Rule
• Sum and Difference Rule
• Product Rule
• Quotient Rule
• Chain Rule

Let $f(x)$ and $g(x)$ be differentiable functions and $k$ be a constant. Then each of the following rules holds:

Sum Rule

The derivative of the sum of a function $f$ and a function $g$ is the same as the sum of the derivative of $f$ and the derivative of $g$.

$
\frac{d}{d x}(f(x)+g(x))=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))
$
In general,

$
\frac{d}{d x}(f(x)+g(x)+h(x)+\ldots \ldots)=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))+\frac{d}{d x}(h(x))+\ldots \ldots
$

Difference Rule
The derivative of the difference of a function $f$ and $a$ function $g$ is the same as the difference of the derivative of $f$ and the derivative of $g$.

$
\begin{aligned}
& \frac{d}{d x}(f(x)-g(x))=\frac{d}{d x}(f(x))-\frac{d}{d x}(g(x)) \\
& \frac{d}{d x}(f(x)-g(x)-h(x)-\ldots \ldots)=\frac{d}{d x}(f(x))-\frac{d}{d x}(g(x))-\frac{d}{d x}(h(x))-\ldots \ldots
\end{aligned}
$

Constant Multiple Rule
The derivative of a constant k multiplied by a function f is the same as the constant multiplied by the derivative of $f$

$
\frac{d}{d x}(k f(x))=k \frac{d}{d x}(f(x))
$

Product rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then,

$
\frac{d}{d x}(f(x) g(x))=g(x) \cdot \frac{d}{d x}(f(x))+f(x) \cdot \frac{d}{d x}(g(x))
$

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Extending the Product Rule

If 3 functions are involved, i.e let $k(x)=f(x) \cdot g(x) \cdot h(x)$
Let us have a function $\mathrm{k}(\mathrm{x})$ as the product of the function $\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})$ and the function $\mathrm{h}(\mathrm{x})$. That is, $k(x)=(f(x) \cdot g(x)) \cdot h(x)$. Thus,

$
k^{\prime}(x)=\frac{d}{d x}(f(x) g(x)) \cdot h(x)+\frac{d}{d x}(h(x)) \cdot(f(x) g(x))
$

[By applying the product rule to the product of $f(x) g(x)$ and $h(x)$.]

$
\begin{aligned}
& =\left(f^{\prime}(x) g(x)+g^{\prime}(x) f(x)\right) h(x)+h^{\prime}(x) f(x) g(x) \\
& =f^{\prime}(x) g(x) h(x)+f(x) g^{\prime}(x) h(x)+f(x) g(x) h^{\prime}(x)
\end{aligned}
$

Quotient Rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then

$
\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\frac{g(x) \cdot \frac{d}{d x}(f(x))-f(x) \cdot \frac{d}{d x}(g(x))}{(g(x))^2}
$
OR
if $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime}(x)=\frac{f^{\prime}(x) g(x)-g^{\prime}(x) f(x)}{(g(x))^2}$
As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives.

Chain Rule:
If $u(x)$ and $v(x)$ are differentiable funcitons, then $u o v(x)$ or $u[v(x)]$ is also differentiab

Chain Rule:

If $u(x)$ and $v(x)$ are differentiable funcitons, then $u o v(x)$ or $u[v(x)]_{\text {is also differentiable. }}$
If $y=u o v(x)=u[v(x)]$, then

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} u\{v(x)\}}{\mathrm{d}\{v(x)\}} \times \frac{\mathrm{d}}{\mathrm{d} x} v(x)
$

is known as the chain rule. Or,

$
\text { If } y=f(u) \text { and } u=g(x) \text {, then } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} u} \cdot \frac{\mathrm{d} u}{\mathrm{~d} x}
$
The chain rule can be extended as follows:
If $y=[\operatorname{uovow}(x)]=u[v\{w(x)\}]$, then

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}[u[v\{w(x)\}]}{\mathrm{d} v\{w(x)\}} \times \frac{\mathrm{d}[v\{w(x)\}]}{\mathrm{d} w(x)} \times \frac{\mathrm{d}[w(x)]}{\mathrm{d} x}
$

Solved Examples Based On Rules of Differentiation:

Example 1: If $y=y(x)$ is the solution of the differential equation, $e^y\left[\frac{d y}{d x}-1\right]=e^x$ such that $y(0)=0$, then $y(1)$ is equal to: [JEE Main 2020]
1) $\log _e 2$
2) $2 e$
3) $2+\log _e 2$
4) $1+\log _e 2$

Solution: $\square$

$
\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{e^y}{e^x}\right)=\frac{e^y \cdot e^x \frac{\mathrm{d} y}{\mathrm{~d} x}-e^x \cdot e^y}{\left(e^x\right)^2} \\
& \int \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{e^y}{e^x}\right)=\int 1 \cdot d x \\
& \left(\frac{e^y}{e^x}\right)=x \\
& \left(e^y\right)=x e^x+c
\end{aligned}
$
Since $y(0)=0$

$
\begin{aligned}
& c=1 \\
& y(1)=1+\ln 2
\end{aligned}
$

Hence, the answer is the option (4).

Example 2: The value of $\log _e 2 \frac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right)$ at $x=\frac{\pi}{4}$ is:
[JEE Main 2022]
1) $-2 \sqrt{2}$
2) $2 \sqrt{2}$
3) -4
4) 4

Solution:

$
\begin{aligned}
& \log 2 \cdot \frac{d}{d x}\left(\frac{\log (\operatorname{cosec} x)}{\log (\cos x)}\right) \\
& =-\log 2 \cdot \frac{d}{d x}\left(\frac{\log (\sin x)}{\log (\cos x)}\right) \\
& =-\log 2 \cdot \frac{\log (\cos x) \cdot \frac{\cos x}{\sin x}-\log (\sin x) \frac{(-\sin x)}{\cos x}}{(\log \cos x)^2}
\end{aligned}
$
At $\mathrm{x}=\frac{\pi}{4}$

$
\begin{aligned}
& =-\log 2 \cdot \frac{\log \left(\frac{1}{\sqrt{2}}\right)+\log \left(\frac{1}{\sqrt{2}}\right)}{\left(\log \left(\frac{1}{\sqrt{2}}\right)\right)^2} \\
& =-\log 2 \cdot \frac{\left(2 \log \left(\frac{1}{\sqrt{2}}\right)\right)}{\left(\log \left(\frac{1}{\sqrt{2}}\right)\right)^2}
\end{aligned}
$
$\begin{aligned} & =\frac{-\log 2 \cdot 2}{\log \left(\frac{1}{\sqrt{2}}\right)} \\ & =\frac{-2 \log 2}{-\log (\sqrt{2})} \\ & =\frac{2 \log 2}{\frac{1}{2} \log 2} \\ & =4\end{aligned}$
Hence, the answer is the option (4).

Example 3: The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha$ has at least one solution in $\left(0, \frac{\pi}{2}\right)$ is
[JEE Main 2021]
1) 7
2) 8
3) 9
4) 10

Solution:
Let $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$

$
y=\frac{4-3 \sin x}{\sin x(1-\sin x)}
$
Let $\sin \mathrm{x}=\mathrm{t}$ when $t \in(0,1)$.

$
y=\frac{4-3 t}{t-t^2}
$

$\begin{aligned} & \frac{d y}{d t}=\frac{-3\left(t-t^2\right)-(1-2 t)(4-3 t)}{\left(t-t^2\right)^2}=0 \\ & \Rightarrow 3 \mathrm{t}^2-3 \mathrm{t}-\left(4-11 \mathrm{t}+6 \mathrm{t}^2\right)=0 \\ & \Rightarrow 3 \mathrm{t}^2-8 \mathrm{t}+4=0 \\ & \Rightarrow 3 \mathrm{t}^2-6 \mathrm{t}-2 \mathrm{t}+4=0 \\ & \Rightarrow \mathrm{t}=\frac{2}{3} \quad \text { and } \quad \mathrm{t} \neq 2 \\ & \frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha \\ & \frac{12}{2}+\frac{3}{3-2}=\alpha \\ & \alpha=9\end{aligned}$

Hence, the answer is the option 3.

Example 4:
Let $f$ and $g$ be differentiable functions on $R$ such that $f o g$ is the identity function. If for some $a, b \in \mathbf{R}, g^{\prime}(a)=5$ and $g(a)=b_{\text {then }} f^{\prime}(b)$ is equal to : [JEE Main 2020]
1) $\frac{2}{5}$
2) 5
3) 1
4) $\frac{1}{5}$

Solution:

Let f and g be functions. For all x in the domain of g for which g is differentiable at x and f is differentiable at $g(x)$, the derivative of the composite function

$
\begin{aligned}
& h(x)=(f \circ g)(x)=f(g(x)) \text { Is given by } \\
& h^{\prime}(x)=f^{\prime}(g(x)) \cdot g^{\prime}(x)
\end{aligned}
$
Composites of Three or More Functions
For all values of $x$ for which the function is differentiable, if $k(x)=h(f(g(x)))$ Then,

$
\begin{aligned}
& k^{\prime}(x)=h^{\prime}(f(g(x))) \cdot f^{\prime}(g(x)) \cdot g^{\prime}(x) \\
& f(g(x))=x \\
& \Rightarrow f^{\prime}(g(x)) \cdot g^{\prime}(x)=1 \\
& P u t x=a \\
& \Rightarrow f^{\prime}(g(a)) g^{\prime}(a)=1 \\
& \Rightarrow f^{\prime}(b) \times 5=1 \Rightarrow f^{\prime}(b)=\frac{1}{5}
\end{aligned}
$

Example 5: If $f(x)=\sin ^{-1}\left(\frac{2 \times 3^x}{1+9^x}\right)$, then $f^{\prime}\left(-\frac{1}{2}\right)$ equals:
[JEE Main 2018]
1) $-\sqrt{3} \log _e \sqrt{3}$
2) $\sqrt{3} \log \sqrt{3}$
3) $-\sqrt{3} \log _e 3$
4) $\sqrt{3} \log _e 3$

Solution:
As we learned,
Chain Rule for differentiation (indirect) -
Let $y=f(x)$ is not in standard form then

$
\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}
$

Now

$\begin{aligned} & f(x)=\sin ^{-1}\left(\frac{2 \times 3^x}{1+9^x}\right) \\ & =2 \tan ^{-1} 3^x \\ & \because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^2} \quad \text { if }-1 \leq x \leq 1 \\ & \frac{d}{d u}(\arctan (u))=\frac{1}{u^2+1} \\ & \frac{d}{d x}\left(3^x\right)=\ln (3) \cdot 3^x \\ & f^{\prime}(x)=2 \times \frac{1}{1+\left(3^x\right)^2} \times 3^x \times \ln 3 \\ & f^{\prime}\left(\frac{-1}{2}\right)=2 \times \frac{1}{1+\left(3^{-1}\right)} \times 3^{\frac{-1}{2}} \times \ln 3 \\ & =2 \times \frac{3}{4} \times \frac{1}{\sqrt{3}} \times \ln 3 \\ & =\sqrt{3} \times \frac{1}{2} \ln 3\end{aligned}$

Hence, the answer is the option 2.

Summary

The differentiation rules help us to evaluate the derivatives of some particular functions, instead of using the general method of differentiation. Some important rules are the sum rule, product rule, chain rule, etc. Differentiation is an important concept of Calculus. It provides a deeper understanding of mathematical ideas paramount for later developments in many scientific and engineering disciplines.