Derivative of Polynomials and Trigonometric Functions

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. $\frac{d}{d x}(\csc (\mathbf{x}))=-\csc (\mathbf{x}) \cot (\mathbf{x})$

Recommended Video Based on Derivatives

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).