Derivative of Polynomials and Trigonometric Functions

Differentiation is one of the important parts of Calculus, which applies in measuring the rate of change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions, the maximum and minimum of functions can be dertermined and problems on motion, growth, and decay can be solved and many more. These concepts of differentiation have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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This Story also Contains

1. Derivative of a Function
2. Derivative of the Trigonometric Functions
3. Solved Examples Based On Derivatives:
4. Summary

In this article, we will cover the concept of Differentiation of Polynomial Functions. 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 five questions have been asked on this concept, including three in 2020, and two in 2023.

Derivative of a Function

Let $f$ be defined on an open interval $I \subseteq R$ containing the point $x$ and suppose that $\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ exists. Then $f$ is said to be differentiable at $x$ and the derivative of $f$ at $x$ denoted by $f'(x)$, given by $f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

Derivative of a Polynomial Function

The derivates for all basic functions are,

$\begin{aligned} & 1 . \quad \frac{d}{d x}(\text { constant })=0 \\ & \begin{aligned} f(x) & =c \text { and } f(x+h)=c \\ f^{\prime}(x) & =\lim\limits _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim\limits _{h \rightarrow 0} \frac{c-c}{h} \\ & =\lim\limits _{h \rightarrow 0} \frac{0}{h} \\ & =\lim\limits _{h \rightarrow 0} 0=0\end{aligned}\end{aligned}$

So the derivative of a constant function is zero. Also since a constant function is a horizontal line the slope of a constant function is 0.

2. $\frac{d}{d x}\left(\mathbf{x}^{\mathbf{n}}\right)=\mathbf{n} \mathbf{x}^{\mathbf{n}-1}$

For $f(x)=x^n$ where $n$ is a positive integer, we have

$
f^{\prime}(x)=\lim\limits _{h \rightarrow 0} \frac{(x+h)^n-x^n}{h}
$
Since,

$
(x+h)^n=x^n+n x^{n-1} h+\binom{n}{2} x^{n-2} h^2+\binom{n}{3} x^{n-3} h^3+\ldots+n x h^{n-1}+h^n
$

we can write

$
(x+h)^n-x^n=n x^{n-1} h+\binom{n}{2} x^{n-2} h^2+\binom{n}{3} x^{n-3} h^3+\ldots+n x h^{n-1}+h^n
$
Next, divide both sides by h:

$
\frac{(x+h)^n-x^n}{h}=\frac{n x^{n-1} h+\binom{n}{2} x^{n-2} h^2+\binom{n}{3} x^{n-3} h^3+\ldots+n x h^{n-1}+h^n}{h}
$

Thus,

$
\frac{(x+h)^n-x^n}{h}=n x^{n-1}+\binom{n}{2} x^{n-2} h+\binom{n}{3} x^{n-3} h^2+\ldots+n x h^{n-2}+h^{n-1}
$
Finally,

$
\begin{aligned}
f^{\prime}(x) & =\lim\limits _{h \rightarrow 0} \frac{(x+h)^n-x^n}{h} \\
& =\lim\limits _{h \rightarrow 0}\left(n x^{n-1}+\binom{n}{2} x^{n-2} h+\binom{n}{3} x^{n-3} h^2+\ldots+n x h^{n-1}+h^n\right) \\
& =n x^{n-1}
\end{aligned}
$

Derivative of Exponential and Logarithmic Functions

3. $\frac{d}{d x}\left(\mathbf{a}^{\mathrm{x}}\right)=\mathbf{a}^{\mathrm{x}} \log _{\mathrm{e}} \mathbf{a}$

$
\begin{array}{rlr}
f(x) & =a^x \text { and } f(x+h)=a^{x+h} \\
f^{\prime}(x) & =\lim\limits _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim\limits _{h \rightarrow 0} \frac{a^{x+h}-a^x}{h} \\
& =a^x \lim\limits _{h \rightarrow 0} \frac{a^h-1}{h} \\
& =a^x \log _e a \quad & \\
& {\left[\because \lim\limits _{x \rightarrow 0} \frac{a^x-1}{x}=\log _e a\right]}
\end{array}
$

4. $\quad \frac{d}{d x}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{e}^{\mathrm{x}} \log _{\mathrm{e}} \mathrm{e}=\mathrm{e}^{\mathrm{x}}$

5. $\frac{d}{d x}\left(\log _{\mathrm{e}}|\mathbf{x}|\right)=\frac{\mathbf{1}}{\mathbf{x}}, \quad \mathbf{x} \neq 0$

$
\begin{aligned}
& f(x)=\log _e(x) \text { and } f(x+h)=\log _e(x+h) \\
& f^{\prime}(x)=\lim\limits _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim\limits _{h \rightarrow 0} \frac{\log _e(x+h)-\log _e(x)}{h} \\
& =\lim\limits _{h \rightarrow 0} \frac{\log _e\left(\frac{x+h}{x}\right)}{h} \\
& =\lim\limits _{h \rightarrow 0} \frac{\log _e\left(1+\frac{h}{x}\right)}{\frac{h}{x} \cdot x} \\
& =\frac{1}{x} \quad\left[\because \lim\limits _{x \rightarrow 0} \frac{\log _e(1+x)}{x}=1\right]
\end{aligned}
$

6. $\quad \frac{d}{d x}\left(\log _{\mathbf{a}}|\mathbf{x}|\right)=\frac{1}{\mathbf{x} \log _{\mathrm{e}} \mathbf{a}}, \quad \mathbf{x} \neq 0$

Derivative of the Trigonometric Functions

The derivatives of the basic trigonometric functions are

Derivative of the Trigonometric Functions (sin/cos/tan)

1. $\frac{d}{d x}(\sin (\mathbf{x}))=\cos (\mathbf{x})$
$f(x)=\sin (x)$ and $f(x+h)=\sin (x+h)$

$
\begin{aligned}
f^{\prime}(x) & =\lim\limits _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim\limits _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x)}{h} \\
& =\lim\limits _{h \rightarrow 0} \frac{2 \sin \frac{h}{2} \cdot \cos \left(\frac{2 x+h}{2}\right)}{h} \\
& =\lim\limits _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \cdot \lim\limits _{h \rightarrow 0} \cos \left(\frac{2 x+h}{2}\right) \\
& =\cos x \quad\left[\because \lim\limits _{x \rightarrow 0} \frac{\sin (x)}{x}=1\right]
\end{aligned}
$

Similarly,

2. $\frac{d}{d x}(\cos (\mathbf{x}))=-\sin (\mathbf{x})$
3. $\frac{d}{d x}(\tan (\mathbf{x}))=\sec ^2(\mathbf{x})$

Derivative of the Trigonometric Functions (cot/sec/csc)

4. $\frac{d}{d x}(\cot (\mathbf{x}))=-\csc ^2(\mathbf{x})$
5. $\frac{d}{d x}(\sec (\mathbf{x}))=\sec (\mathbf{x}) \tan (\mathbf{x})$
6. $\frac{d}{d x}(\csc (\mathbf{x}))=-\csc (\mathbf{x}) \cot (\mathbf{x})$

Solved Examples Based On Derivatives:

Example 1: If $f^{\prime}(x)=\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{2}<x<\frac{\pi}{2}$, and $f(0)=0$, then $f(1)$ is equal to: [JEE Main 2020]
1) $\frac{\pi+1}{4}$
2) $\frac{\pi+2}{4}$
3) $\frac{1}{4}$
4) $\frac{\pi-1}{4}$:

Solution:

$
\begin{aligned}
& f^{\prime}(x)=\tan ^{-1}(\sec x+\tan x)=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) \\
& \Rightarrow \tan ^{-1}\left(\frac{1-\cos \left(\frac{\pi}{2}+x\right)}{\sin \left(\frac{\pi}{2}+x\right)}\right)=\tan ^{-1}\left(\frac{2 \sin ^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}\right) \\
& =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right)=\frac{\pi}{4}+\frac{x}{2} \\
& \left(f^{\prime}(x)\right)=\frac{\pi}{4}+\frac{x}{2} \\
& f(x)=\frac{\pi}{4} x+\frac{x^2}{4}+c \\
& \quad f(0)=c=0 \Rightarrow \quad f(x)=\frac{\pi}{4} x+\frac{x^2}{4}
\end{aligned}
$
So $f(1)=\frac{\pi+1}{4}$

Hence, the answer is the option 1.

Example 2: Let $x^k+y^k=a^k,(a, k>0)$ and $\frac{d y}{d x}+\left[\frac{y}{x}\right]^{\frac{1}{3}}=0$, then $k$ is: [JEE Main 2020]
1) $\frac{1}{3}$
2) $\frac{3}{2}$
3) $\frac{2}{3}$
4) $\frac{4}{3}$

Solution:

$\begin{aligned} & x^k+y^k=a^k \quad(a, k>0) \\ & \frac{d y}{d x}+\left(\frac{y}{x}\right)^{1 / 3}=0 \\ & \Rightarrow k x^{k-1}+k y^{k-1} \cdot \frac{d y}{d x}=0\end{aligned}$

$\begin{aligned} & \frac{d y}{d x}+\left(\frac{x}{y}\right)^{k-1}=0 \\ & k-1=-1 / 3 \\ & k=2 / 3\end{aligned}$

Hence, the answer is the option (3).

Example 3: Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be a function of x satisfying $y \sqrt{1-x^2}=k-x \sqrt{1-y^2}$, where k is a constant and $y\left(\frac{1}{2}\right)=-\frac{1}{4} . \quad \frac{d y}{d x}$ at $x=\frac{1}{2}$, is equal to:
[JEE Main 2020]
1) $-\frac{\sqrt{5}}{2}$
2) $\frac{\sqrt{5}}{2}$
3) $-\frac{\sqrt{5}}{4}$
4) $\frac{2}{\sqrt{5}}$

Solution:

Here,

$\begin{aligned} & x=1 / 2, y=-1 / 4 \Rightarrow x y=-1 / 8 \\ & y \cdot \frac{1 \cdot(-2 x)}{2 \sqrt{1-x^2}}+y^{\prime} \sqrt{1-x^2}=-\left[1 \cdot \sqrt{\left(1-y^2\right)+\frac{x \cdot(-2 y)}{2 \cdot \sqrt{1-y^2}} y^{\prime}}\right] \\ & -\frac{x y}{\sqrt{1-x^2}}+y^{\prime} \sqrt{1-x^2}=-\sqrt{1-y^2}+\frac{x y \cdot y^{\prime}}{\sqrt{1-y^2}} \\ & y^{\prime}\left(\sqrt{1-x^2}-\frac{x y}{\sqrt{1-y^2}}\right)=\frac{x y}{\sqrt{1-x^2}}-\sqrt{1-y^2} \\ & \text { put } x=1 / 2 \text { and } y=-1 / 4 \\ & \text { we get } y^{\prime}=-\frac{\sqrt{5}}{2}\end{aligned}$

Hence, the answer is the option 1.

Example 4: Let

$
f(x)=\sum_{k=1}^{10} k x^k, x \in R \text { If } 2 f(2)-f^{\prime}(2)=119(2)^n+1
$

then $n$ is equal to
[JEE Main 2023]
1) 10
2) -1
3) 2
4) 4

Solution:
$
\begin{aligned}
& f(x)=\sum_{k=1}^{10} k x^k \\
& \quad \Rightarrow f(x)=x+2 x^2+3 x^3+\cdots+9 x^9+10 x^{10}-(i) \\
& x f(x)=x^2+2 x^3+\ldots+9 x^{10}+10 x^{11} \ldots \text { (ii) }
\end{aligned}
$

"(i) - (ii)"

$
\begin{aligned}
& f(x)(1-x)=x+x^2+x^3+\cdots+x^{10}-10 x^{11} \\
& f(x)(1-x)=\frac{x\left(1-x^{10}\right)}{1-x}-10 x^{11} \\
& f(x)=\frac{x\left(1-x^{10}\right)}{(1-x)^2}-\frac{10 x^{11}}{(1-x)}
\end{aligned}
$
$
\begin{aligned}
& f(2)=2+g(2)^{11} \\
& (1-x)^2 f(x)=x\left(1-x^{10}\right)-10 x^{11}(1-x)
\end{aligned}
$

diff. w.r.t. $x$

$
\begin{aligned}
& (1-x)^2 f^{\prime}(2)+f(2) 2(1-x)(-1) \\
& =x\left(-10 x^9\right)+\left(1-x^{10}\right)-10 x^{11}(-1)-(1-x)(110) x^{10}
\end{aligned}
$

put $x=2$

$
\begin{aligned}
& \mathrm{f}^{\prime}(2)+\mathrm{f}(2)(2)=-10(2)^{10}+1-2^{10}+10(2)^{11}-110(2)^{10}+110(2)^{11} \\
& =(-121) 2^{10}+(120) 2^{11}+1 \\
& =2^{10}(240-121)+1 \\
& =119(2)^{10}+1 \\
& \mathrm{n}=10
\end{aligned}
$

Hence, the answer is 10.

Example 5: If $f(x)=x^2+g^{\prime}(1) x+g^{\prime \prime}(2)$ and $g(x)=f(1) x^2+x f^{\prime}(x)+f^{\prime \prime}(x)$, then the value of $f(4)-g(4)$ is equal to
[JEE Main 2023]
1) 14
2) -1
3) 2
4) 4

Solution:

$
\begin{aligned}
& \text { let } g^{\prime}(1)=\mathrm{A} \\
& g^{\prime \prime}(2)=\mathrm{B} \\
& f(\mathrm{x})=\mathrm{x}^2+\mathrm{Ax}+\mathrm{B} \\
& f(1)=\mathrm{A}+\mathrm{B}+1 \\
& f^{\prime}(\mathrm{x})=2 \mathrm{x}+\mathrm{A} \\
& f^{\prime \prime}(\mathrm{x})=2
\end{aligned}
$

$\begin{aligned} & g(x)=(A+B+1) x^2+x(2 x+A)+2 \\ & \Rightarrow g(x)=x^2(A+B+2)+A x+2 \\ & g^{\prime}(\mathrm{x})=2 \mathrm{x}(\mathrm{A}+\mathrm{B}+2)+\mathrm{A} \\ & g^{\prime}(1)=A \\ & \Rightarrow 2(\mathrm{~A}+\mathrm{B}+2)+A=A \\ & \mathrm{~A}+\mathrm{B}=-2\end{aligned}$
$
\begin{aligned}
& g(x)=2(A+B+2) \\
& g(2)=B \\
& \Rightarrow 2(A+B+2)=B \\
& \Rightarrow 2 A+B=-4
\end{aligned}
$
From (i) and (ii)

$
\begin{aligned}
& \mathrm{A}=-2 \text { and } \mathrm{B}=0 \\
& f(\mathrm{x})=\mathrm{x}^2-2 \mathrm{x} \\
& f(4)=16-8=8 \\
& \mathrm{~g}(\mathrm{x})=-2 \mathrm{x}+2 \\
& \mathrm{~g}(4)=-8+2=-6 \\
& f(4)-g(4)=8-(-6)=14
\end{aligned}
$

Hence, the answer is the option (1).

Summary

Differentiation is an important concept of Calculus. Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point. The differentiation of a constant is zero. With the help of differentiation, we can find the rate of change of one quantity for another. The concept of differentiation is the cornerstone on which the development of calculus rests.