Written By Ritu_Kumari
Last Modified 25-01-2023

Derivatives of Polynomials and Trigonometric Functions

Derivatives of Polynomial and Trigonometric Functions: We use the concept of derivatives to express the rate of change in any function (polynomial function, trigonometric, and inverse trigonometric functions). This considers even the infinitesimally small changes in the dependent variable with respect to small changes in the independent variables i.e. if there is a very small change in the dependent variable with respect to the independent variable that change will also be considered.

The process of finding the derivative is known as differentiation. Differentiation has been used across fields like Science, Engineering, and Physics. It is used to find the rate of change of a quantity, the approximation value, the equation of tangent and normal to a curve, and the minimum and maximum values of algebraic expressions.

Derivatives of a Function

If \(f\) is a real-valued function, then the function defined by \(\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h}\) wherever the limit exists, is defined to be the derivative of \(f\) at \(x\). It is denoted as \(f’\left( x \right)\). Therefore, \(f’\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h}\), provided the limit exists.This method of finding the derivatives is also known as the derivative of a function from the first principle. Here \(f’\left( x \right) = \frac{d}{{dx}}\left( {f\left( x \right)} \right)\) if \(y = f\left( x \right)\) then it is denoted by \(\frac{{dy}}{{dx}}\), i.e., derivative of \(f\left( x \right)\) or \(y\) w.r.t \(x\).

Let’s Learn About Polynomials Here

Important Formulae on Derivatives of a Function

\(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n – 1}}\)
\(\frac{d}{{dx}}\left[ {{{\left( {ax + b} \right)}^n}} \right] = na{\left( {ax + b} \right)^{n – 1}}\)
Derivative of a constant is zero

Points to Remember

For the derivative of \(f\) at a point \(x\) to exist, it is necessary that the function \(f\) is defined in a neighbourhood of the point \(x\), including the point \(x\) itself.
The derived function \(f’\left( x \right)\) is also called the derivative or differential coefficient of \(f\left( x \right)\) with respect to \(x\). If \(f’\left( x \right)\) exists for all \(x\) in \((a,b)\) then \(f\left( x \right)\) is said to be differentiable on \((a,b)\).
A function \(f\) may not be differentiable at each point of its domain \(D\).
The set of those points of \(D\) at which \(f’\left( x \right)\) exists is called the domain of \(f’\).
If a function \(f\) is not continuous at a certain point \(x=c\), then the function \(f\) is not derivable at that point. The converse of this statement is not true in all cases.
Some functions are continuous but do not have a derivative. For example: \(\left| x \right|\) is continuous at \(x=0\), but not differentiable at \(x=0\).
Existence of the derivative on an interval:
a. A real-valued function \(y = f\left( x \right)\) is said to be derivable in \(\left[ {a,b} \right]\) if it is differentiable in \((a,b)\) and R.H.D at \(x=a\) and L.H.D. at \(x=b\), i.e. \(f’\left( {{a^ + }} \right)\) and \(f’\left( {{b^ – }} \right)\) both exist.
Identify the non-derivative of a function at \(x=a\) on a graph:
a. Discontinuity of function
b. Sharp points on the graph
c. Vertical tangent
d. High-frequency oscillation in the neighbourhood

Derivatives of Polynomials

An expression of the form \({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ….. + {a_n}{x^n}\left( {{a_n} \ne 0} \right)\) where \({a_0},{a_1},{a_2},{a_3},….{a_n}\) are real constant is known as a polynomial.
Here, \(n\) is the degree of the polynomial. This is also known as the \({n^{th}}\) degree polynomial.

Theorem: If \(f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ….. + {a_n}{x^n}\left( {{a_n} \ne 0} \right)\), where \({a_0},{a_1},{a_2},{a_3},….{a_n}\) are real constants, then \(f’\left( x \right) = {a_1} + 2{a_2}x + 3{a_3}{x^2} + ….. + n{a_n}{x^{n – 1}}\), i.e. the derivative of a polynomial of degree \(n\) is a polynomial of degree \(n-1\).
Proof: Let \(f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ….. + {a_n}{x^n}\left( {{a_n} \ne 0} \right)\)
Therefore, \(f’\left( x \right) = \frac{d}{{dx}}\left( {{a_0}} \right) + {a_1}\frac{d}{{dx}}\left( x \right) + {a_2}\frac{d}{{dx}}\left( {{x^2}} \right) + {a_3}\frac{d}{{dx}}\left( {{x^3}} \right) + …. + {a_n}\frac{d}{{dx}}\left( {{x^n}} \right)\)
We know that, \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n – 1}}\) and \(\frac{d}{{dx}}\left( c \right) = 0\), where \(c\) is a constant.
\( \Rightarrow f’\left( x \right) = 0 + {a_1}\left( 1 \right) + {a_2}\left( {2x} \right) + {a_3}\left( {3{x^2}} \right) + ….. + {a_n}\left( {n{x^{n – 1}}} \right)\)
\( = {a_1} + 2{a_2}x + 3{a_3}{x^2} + …. + n{a_n}{x^{n – 1}}\)
Note that this is a polynomial of degree \(n-1\).

If \(f\left( x \right)\) and \(g\left( x \right)\) are two differentiable polynomials, then:
i. \(\frac{d}{{dx}}\left( {cf\left( x \right)} \right) = cf’\left( x \right)\) where \(c\) is a constant.
ii. \(\frac{d}{{dx}}\left[ {f\left( x \right) \pm g\left( x \right)} \right] = f’\left( x \right) \pm g’\left( x \right)\)
iii. \(\frac{d}{{dx}}\left[ {f\left( x \right) \times g\left( x \right)} \right] = f’\left( x \right)g\left( x \right) + g’\left( x \right)f\left( x \right)\)
iv. \(\frac{d}{{dx}}\left[ {f\left( x \right)/g\left( x \right)} \right] = \left[ {g\left( x \right)f’\left( x \right) – f\left( x \right)g’\left( x \right)} \right]/{\left( {g\left( x \right)} \right)^2}\) where \(g\left( x \right) \ne 0\)

Example: Find the derivative of \(f\left( x \right) = 7{x^5} + 4{x^3} – 20x\) and the degree of \(f’\left( x \right)\).
Solution: Given, \(f\left( x \right) = 7{x^5} + 4{x^3} – 20x\)
Therefore, \(f’\left( x \right) = \frac{d}{{dx}}\left( {7{x^5}} \right) + \frac{d}{{dx}}\left( {4{x^3}} \right) – \frac{d}{{dx}}\left( {20x} \right)\)
\( = 35{x^4} + 12{x^2} – 20\)
The highest power of \(x\) in \(f’\left( x \right)\) is \(4\).
Therefore, the degree of \(f’\left( x \right)\) is \(4\).

Derivatives of Trigonometric Functions

The trigonometric ratios are \(\sin \,x,\,\cos \,x,\,\tan \,x,\,\sec \,x,\,{\rm{cosec}}\,x\), and \(\cot \,x\).
Derivative of \(\sin \,x\)
Let \(f\left( x \right) = \sin \,x\). Then, by definition, we have
\(f’\left( x \right) = \frac{d}{{dx}}\left( {f\left( x \right)} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {x + h} \right) – \sin \,\,x}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{2\cos \left( {\frac{{2x + h}}{2}} \right)\sin \,\left( {\frac{h}{2}} \right)}}{h}\) \(\,\left[ {\because \,\sin \,x – \,\sin \,y = 2\,\cos \left( {\frac{{x + y}}{2}} \right)\,\sin \left( {\frac{{x – y}}{2}} \right)} \right]\)
\( = \mathop {\lim }\limits_{h \to 0} \,\cos \,\left( {x + \frac{h}{2}} \right)\, = \mathop {\lim }\limits_{h \to 0} \,\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{As}}\,h \to 0,\,\frac{h}{2} \to 0\,} \right]\)
\( = \cos \,x \times 1\,\,\,\,\,\,\left[ {\because \mathop {\lim }\limits_{x \to 0} \,\frac{{\sin \,x}}{x} = 1} \right]\)
\( = \,\cos \,x\)
Hence, \(\frac{d}{{dx}}\left( {\sin \,x} \right) = \cos \,x\)

Derivative of \(\cos \,x\)
Let \(f\left( x \right) = \,\cos \,x\). Then, by definition, we have
\(f’\left( x \right) = \frac{d}{{dx}}\left( {f\left( x \right)} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( {x + h} \right) – \cos \,\,x}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{ – 2\,\sin \left( {\frac{{2x + h}}{2}} \right)\sin \,\left( {\frac{h}{2}} \right)}}{h}\,\,\,\,\,\,\left[ {\because \,\cos \,x – \cos \,y = – 2\,\sin \left( {\frac{{x + y}}{2}} \right)\,\sin \,\left( {\frac{{x – y}}{2}} \right)} \right]\)
\( = \mathop {\lim }\limits_{h \to 0} \, – \sin \,\left( {x + \frac{h}{2}} \right)\, \mathop {\lim }\limits_{h \to 0} \,\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{As}}\,h \to 0,\,\frac{h}{2} \to 0\,} \right]\)
\( = – \sin \,x \times 1\) \(\left[ {\because \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right]\)
\( = – \sin \,x\)
Hence, \(\frac{d}{{dx}}\left( {\cos \,x} \right) = – \sin \,x\)

Derivative of \(\tan \,x\)
Let \(f\left( x \right) = \tan \,x\). Then, by definition, we have
\(f’\left( x \right) = \frac{d}{{dx}}\left( {f\left( x \right)} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\tan \left( {x + h} \right) – \tan \,\,x}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{\sin \,\left( {x + h} \right)}}{{\cos \,\left( {x + h} \right)}} – \frac{{\sin \,x}}{{\cos \,x}}}}{h}\,\,\,\,\,\,\left[ {\because \,\,\tan \,x\, = \frac{{\sin \,x}}{{\cos \,x}}} \right]\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sin \,\left( {x + h} \right)\,\cos \,x – \sin \,x\,\cos \left( {x + h} \right)} \right]}}{{h\,\cos \,\left( {x + h} \right)\,\cos \,x}}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{\left[ {\sin \,\left( {2x + h} \right) + \sin \,h} \right]}}{2} – \frac{{\left[ {\sin \,\left( {2x + h – \sin \,h} \right)} \right]}}{2}}}{{h\,\cos \,\left( {x + h} \right)\,\cos \,x}}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,\left( h \right)}}{{h\,\cos \,\left( {x + h} \right)\,\cos \,x}}\)
\( = \mathop {\lim }\limits_{h \to 0} \,\frac{1}{{\cos \,\left( {x + h} \right)\,\cos \,x}} \times \mathop {\lim }\limits_{h \to 0} \,\frac{{\sin \,(h)}}{h}\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{As}}\,h \to 0,\,\frac{h}{2} \to 0,\mathop {\lim }\limits_{x \to 0} \,\frac{{\sin \,x}}{x} = 1\,} \right]\)
\( = 1 \times \frac{1}{{\cos \,x \times \,\cos \,x}}\)
\( = \frac{1}{{{{\cos }^2}\,x}} = {\sec ^2}\,x\,\,\,\,\,\,\,\left[ {\because \,\sec \,x = \frac{1}{{\cos \,x}}} \right]\)
Hence, \(\frac{d}{{dx}}\left( {\tan \,x} \right) = {\sec ^2}\,x\)

Similarly, we can find the derivative of the other trigonometric functions as well by using the first principle.

Derivatives of all Trigonometric Functions

\(\frac{d}{{dx}}\left( {\sin \,x} \right) = \cos \,x\)
\(\frac{d}{{dx}}\left( {\cos \,x} \right) = – \sin \,x\)
\(\frac{d}{{dx}}\left( {\tan \,x} \right) = {\sec ^2}\,x\)
\(\frac{d}{{dx}}\left( {\cot \,x} \right) = – {\rm{cosec}}^2\,x\)
\(\frac{d}{{dx}}\left( {\sec \,x} \right) = \sec \,x\,\tan \,x\)
\(\frac{d}{{dx}}\left( {{\rm{cosec}}\,x} \right) = – {\rm{cosec}}\,x\,\cot \,x\)

Example: Find the derivatives of \(5\,\sec \,x – 4\,\cos \,x\) w.r.t. \(x\).
Solution: \(f\left( x \right) = 5\,\sec \,x – 4\,\cos \,x\)
Then, \(f’\left( x \right) = 5\,\frac{d}{{dx}}\left( {\sec \,x} \right) – 4\,\frac{d}{{dx}}\left( {\cos \,x} \right)\)
\( = 5\,\sec x\tan x + 4\,\sin x\,\left[ {\because \,\frac{d}{{dx}}\left( {\sec \,x} \right) = \sec \,x\,\tan \,x,\,\frac{d}{{dx}}\left( {\cos \,x} \right) = – \sin \,x\,} \right]\)

Inverse Trigonometric Functions

We are familiar with the trigonometric function. The equation \(\sin \,y = x\) means that \(y\) is an angle whose sine is \(x\).

We can rewrite this statement as \(y = {\sin ^{ – 1}}x\) and this is read as, “\(y\) is equal to sine inverse \(x\)”. Thus, \({\sin ^{ – 1}}x\) is defined as an angle whose sine is \(x\).

Here are the six inverse trigonometric functions with their principal values branches:

Functions	Domain	Range (Principal Value Branch)
\(y = {\sin ^{ – 1}}x\)	\(\left[ { – 1,\,1} \right]\)	\(\left[ { – \frac{\pi }{2},\,\frac{\pi }{2}} \right]\)
\(y = {\cos ^{ – 1}}x\)	\(\left[ { – 1,\,1} \right]\)	\(\left[ {0,\,\pi } \right]\)
\(y = {\tan ^{ – 1}}x\)	\(\mathbb{R}\)	\(\left( { – \frac{\pi }{2},\,\frac{\pi }{2}} \right)\)
\(y = {\cot ^{ – 1}}x\)	\(\mathbb{R}\)	\(\left( {0,\,\pi } \right)\)
\(y = {\sec ^{ – 1}}x\)	\(\mathbb{R} – \left( { – 1,\,1} \right)\)	\(\left[ {0,\,\pi } \right] – \left\{ {\frac{\pi }{2}} \right\}\)
\(y = {\rm{cosec}}^{ – 1}x\)	\(\mathbb{R} – \left( { – 1,\,1} \right)\)	\(\left[ { – \frac{\pi }{2},\,\frac{\pi }{2}} \right] – \left\{ 0 \right\}\)

Here are a few important notes related to inverse trigonometric function, which we will use further in this article:
(i) \(\mathop {\lim }\limits_{x \to 0} \frac{{\,{{\sin }^{ – 1}}x}}{{\,x}} = 1\)
(ii) \(\mathop {\lim }\limits_{x \to 0} \frac{{\,{{\tan }^{ – 1}}x}}{{\,x}} = 1\)
Differential Coefficient of \({\sin ^{ – 1}}\,x\)
Let \(y = {\sin ^{ – 1}}\,x\), where \(x \in \left( { – 1,\,1} \right)\)
\( \Rightarrow x = \sin y\,\,\,…….\left( i \right)\)
Let \(\delta y\) be a small increment in \(y\) and \(\delta x\) the corresponding increment in \(x\)
\(\therefore x + \delta x = \sin \,\left( {y + \delta y} \right)\,\,\,\,……\left( {ii} \right)\)
Subtracting \(i\) from \(ii\), we get
\(\delta x = \sin \,\left( {y + \delta y} \right) – \sin \,y = 2\,\cos \left( {y + \frac{{\delta y}}{2}} \right) \cdot \sin \frac{{\delta y}}{2}\,\,\,….\left( {iii} \right)\)
Dividing both sides of \((iii)\) by \(\delta y\), we get
\(\frac{{\delta x}}{{\delta y}} = \frac{2}{{\delta y}}\cos \left( {y + \frac{{\delta y}}{2}} \right)\, \cdot \,\sin \frac{{\delta y}}{2} = \cos \left( {y + \frac{{\delta y}}{2}} \right) \cdot \frac{{\sin \frac{{\delta y}}{2}}}{{\frac{{\delta y}}{2}}}\)
Proceeding to limits as \(\delta y \to 0\), we get
\(\mathop {\lim }\limits_{\delta y \to 0} \frac{{\delta x}}{{\delta y}} = \mathop {\lim }\limits_{\delta y \to 0} \,\cos \,\left( {y + \frac{{\delta y}}{2}} \right) \cdot \,\mathop {\lim }\limits_{\delta y \to 0} \frac{{\sin \frac{{\delta y}}{2}}}{{\frac{{\delta y}}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{As}}\,\delta y \to 0,\,\frac{{\delta y}}{2} \to 0} \right]\)
\(\therefore \,\,\,\frac{{dx}}{{dy}} = \cos y \cdot 1 = \cos \,y\,\,\,\,\left[ {\because \,\mathop {\lim }\limits_{x \to 0} \,\frac{{{{\sin }^{ – 1}}\,x}}{x} = 1\,} \right]\)
i.e., \(\frac{{dy}}{{dx}} = \frac{1}{{\cos \,y}} = \frac{1}{{ \pm \sqrt {1 – {{\sin }^2}\,y} }} = \pm \frac{1}{{\sqrt {1 – {x^2}} }}\)
Taking the principal value of \(y = {\sin ^{ – 1}}x\) which lies between \( – \frac{\pi }{2}\) and \( \frac{\pi }{2}\); \(cos y\) is positive.
\(\therefore \,\) Rejecting the negative sign, we get \(\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 – {x^2}} }}\)
Hence, \(\frac{d}{{dx}}\left( {{{\sin }^{ – 1}}x} \right) = \frac{1}{{\sqrt {1 – {x^2}} }},\,x \in \left( { – 1,\,1} \right)\),

Similarly, we can find derivatives of other inverse trigonometric functions.

Inverse Trigonometric Functions	Derivatives
\(y = {\sin ^{ – 1}}x\)	\(\frac{1}{{\sqrt {1 – {x^2}} }},\,x \in \left( { – 1,\,1} \right)\)
\(y = {\cos ^{ – 1}}x\)	\( – \frac{1}{{\sqrt {1 – {x^2}} }},\,x \in \left( { – 1,\,1} \right)\)
\(y = {\tan ^{ – 1}}x\)	\(\frac{1}{{1 + {x^2}}},\,x \in \mathbb{R}\)
\(y = {\cot ^{ – 1}}x\)	\( – \frac{1}{{1 + {x^2}}},\,x \in \mathbb{R}\)
\(y = {\sec ^{ – 1}}x\)	\(\frac{1}{{x\sqrt {{x^2} – 1} }}\)
\(y = {\rm{cosec}}^{ – 1}x\)	\(-\frac{1}{{x\sqrt {{x^2} – 1} }}\)

Solved Examples

Q.1. Find the derivative of \(f\left( x \right) = {\left( {ax + b} \right)^n}\), for any positive integer \(n\).
Sol:
Let \(f\left( x \right) = {\left( {ax + b} \right)^n}\), where \(n\) is any positive integer.
By definition, we have
\(f’\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – x}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left[ {a\left( {x + h} \right) + b} \right]}^n} – {{\left( {ax + b} \right)}^n}}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left[ {\left( {ax + b} \right) + ah} \right]}^n} – {{\left( {ax + b} \right)}^n}}}{h}\)
By the binomial theorem, we have
\({\left[ {\left( {ax + b} \right) + ah} \right]^n}{ = ^n}{c_0}{\left( {ax + b} \right)^n}{ + ^n}{c_1}{\left( {ax + b} \right)^{n – 1}} \cdot \left( {ah} \right){ + ^n}{c_2}{\left( {ax + b} \right)^{n – 2}} \cdot {\left( {ah} \right)^2} + ….{ + ^n}{c_n}{\left( {ah} \right)^n}\)
\(\therefore \,{\left[ {\left( {ax + b} \right) + ah} \right]^n} – {\left( {ax + b} \right)^n} = h\left[ {na{{\left( {ax + b} \right)}^{n – 1}}{ + ^n}{c_2}{{\left( {ax + b} \right)}^{n – 2}} \cdot {a^2}h + ….{ + ^n}{c_n}{a^n}{h^{n – 1}}} \right]\)
Then, we get
\(f’\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{h\left[ {na{{\left( {ax + b} \right)}^{n – 1}}{ + ^n}{c_2}{{\left( {ax + b} \right)}^{n – 2}} \cdot {a^2}h + ….{ + ^n}{c_n}{a^n}{h^{n – 1}}} \right]}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \left[ {na{{\left( {ax + b} \right)}^{n – 1}}{ + ^n}{c_2}{{\left( {ax + b} \right)}^{n – 2}} \cdot {a^2}h + ….{ + ^n}{c_n}{a^n}{h^{n – 1}}} \right]\)
\( = na{\left( {ax + b} \right)^{n – 1}}\)
Hence, \(\frac{d}{{dx}}{\left( {ax + b} \right)^n} = na{\left( {ax + b} \right)^{n – 1}}\).

Q.2. Find the derivative of \(f\left( x \right) = 8{x^4} + 4{x^3} – 2{x^2} – 10x + 6\), and hence find \(f’\left( 1 \right)\).
Sol:
Given: \(f\left( x \right) = 8{x^4} + 4{x^3} – 2{x^2} – 10x + 6\)
Therefore, \(\begin{gathered} f’\left( x \right) = \frac{d}{{dx}}\left( {8{x^4}} \right) + \frac{d}{{dx}}\left( {4{x^3}} \right) – \frac{d}{{dx}}\left( {2{x^2}} \right) – \frac{d}{{dx}}\left( {10x} \right) + \frac{d}{{dx}}\left( 6 \right) \hfill \\ \hfill \\\end{gathered} \)
\( = 32{x^3} + 12{x^2} – 4x – 10\)
Now, \(f’\left( 1 \right) = 32{\left( 1 \right)^3} + 12{\left( 1 \right)^2} – 4\left( 1 \right) – 10 = 30\)
Therefore, \(f’\left( 1 \right)\) is \(30\).

Q.3. Find the derivative of the function \(\sec \,2x\tan \,2x\) with respect to \(x\).
Sol:
Let \(f\left( x \right) = \sec \,2x\tan \,2x\)
\(\therefore f’\left( x \right) = \frac{d}{{dx}}\left( {\sec \,2x\tan \,2x} \right)\)
\( = \,\sec \,2x \cdot \frac{d}{{dx}}\left( {\tan \,2x} \right) + \tan \,2x \cdot \frac{d}{{dx}}\left( {\sec \,2x} \right)\)
\( = \,\sec \,2x \cdot {\sec ^2}\,2x \cdot 2 + \tan \,2x \cdot \left( {\sec \,2x \cdot \tan \,2x} \right) \cdot 2\)
\( = 2\sec \,2x\left( {{{\sec }^2}\,2x + {{\tan }^2}\,2x} \right)\)
\( = 2\sec \,2x\left( {1 + 2{{\tan }^2}\,2x} \right)\)

Q.4. Find the derivative if it exist for the given function
\(f\left( x \right) = \left\{ \begin{gathered} 3{x^2} + 1,\,x \geqslant 0 \hfill \\ – {x^3},\,\,\,\,\,\,\,x < 0 \hfill \\\end{gathered} \right.\)
Sol:
Given function \(f\left( x \right) = \left\{ \begin{gathered} 3{x^2} + 1,\,x \geqslant 0 \hfill \\ – {x^3},\,\,\,\,\,\,\,x < 0 \hfill \\\end{gathered} \right.\)
If we direct differentiate the given function, we get
\(f’\left( x \right) = \left\{ \begin{gathered}6x,\,\,\,x \geqslant 0 \hfill \\ – 3{x^3},\,\,\,\,\,\,\,x < 0 \hfill \\\end{gathered} \right.\)
Hence, \(Lf’\left( 0 \right) = Rf’\left( 0 \right) = 0\)
\( \Rightarrow f’\left( 0 \right) = 0\) but it is not correct.
Since \(f\left( x \right)\) is not continuous at \(x=0\), we cannot proceed for differentiability. This can take place when the function is a piece-wise function. Let us evaluate both L.H.D. & R.H.D. by the first principle.
\(Lf’\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ – }} \frac{{3{x^2} + 1}}{x} = – \infty \)
\(Rf’\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^3}}}{x} = 0\)
Therefore, \(f\) is not differentiable at \(x=0\).
Hence, the derivative of \(f\) does not exist at \(x=0\).

Q.5. If \(y = {\cos ^{ – 1}}\left( {2{x^2} – 1} \right)\) find \(\frac{{dy}}{{dx}}\).
Sol:
The given function is an inverse trigonometric function. So, we use the substitution method to find the derivative of the function.
Given: \(y = {\cos ^{ – 1}}\left( {2{x^2} – 1} \right)\)
Put \(x = \cos \,\theta \) so that \(\theta = {\cos ^{ – 1}}x\)
\(y = {\cos ^{ – 1}}\left( {2{{\cos }^2}\,\theta – 1} \right)\)
\(y = {\cos ^{ – 1}}\left( {\cos \,2\theta } \right) = 2\theta = 2{\cos ^{ – 1}}x\)
\(\therefore \,\frac{{dy}}{{dx}} = – \frac{2}{{\sqrt {1 – {x^2}} }}\)
Hence, the derivative of \(y = {\cos ^{ – 1}}\left( {2{x^2} – 1} \right)\) is \( – \frac{2}{{\sqrt {1 – {x^2}} }}\)

Summary

If \(f\) is a real-valued function, then the function defined by \(\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h}\). Wherever the limit exists, it is defined to be the derivative of \(f\) at \(x\) and is denoted by \(f’\left( x \right)\). If \(f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ….. + {a_n}{x^n}\left( {{a_n} \ne 0} \right)\), where \({a_0},{a_1},{a_2},{a_3},….{a_n}\) are real constant, then \(f’\left( x \right) = {a_1} + 2{a_2}x + 3{a_3}{x^2} + ….. + n{a_n}{x^{n – 1}}\), i.e. the derivative of a polynomial of degree \(n\) is a polynomial of degree \(n-1\). If \(f\left( x \right)\) and \(g\left( x \right)\) are two differentiable functions, then addition, subtraction, multiplication, and division of polynomials are also differentiable. This can be done by using suitable formulas of differentiation. Derivatives of trigonometric functions and inverse trigonometric functions also exist in their domain.

Frequently Asked Questions (FAQs)

Q.1. What is the derivative of a polynomial function?
Ans: The derivative of a polynomial of degree \(n\) is a polynomial of degree \(n-1\).i.e., the derivative of a polynomial function is also a polynomial function with degree \(n-1\). The derivative of the polynomial function \(f\left( x \right)\) is denoted as \(f’\left( x \right)\).

Q.2. How do you derive derivatives of trigonometric functions?
Ans: The derivative of a trigonometric function can be done using the first principal of differentiation or using the formulas given below.
1. \(\frac{d}{{dx}}\left( {\sin \,x} \right) = \cos \,x\)
2. \(\frac{d}{{dx}}\left( {\cos \,x} \right) = – \sin \,x\)
3. \(\frac{d}{{dx}}\left( {\tan \,x} \right) = {\sec ^2}\,x\)
4. \(\frac{d}{{dx}}\left( {\cot \,x} \right) = – {\rm{cosec}}^2\,x\)
5. \(\frac{d}{{dx}}\left( {\sec \,x} \right) = \sec \,x\,\tan \,x\)
6. \(\frac{d}{{dx}}\left( {{\rm{cosec}}\,x} \right) = – {\rm{cosec}}\,x\,\cot \,x\)

Q.3. What is the meaning of the derivative of a trigonometric function?
Ans: As per the definition of differentiation, the derivative of a trigonometric function is nothing but the rate of change of the trigonometric function with respect to the rate of change of an angle (i.e. independent variable).
For example: \(\frac{d}{{dx}}\sin \left( a \right) = \cos \,a\) is nothing but the rate of change of \(\sin \,x\) at a particular angle \(x=a\) is given by the cosine of that angle.

Q.4. What is the derivative of \(4\)?
Ans: Since \(4\) is constant with respect to \(x\), we know that derivative of a constant is zero.
Therefore, the derivative of \(4\) with respect to \(x\) is \(0\).

Q.5. What is the derivative formula?
Ans: If \(f\) is a real-valued function, then the function defined by
\(\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h}\)
Wherever the limit exists, it is defined to be the derivative of \(f\) at \(x\) and is denoted by \(f’\left( x \right)\).
Therefore, \(f’\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h}\), provided the limit exists.

NCERT Solutions for Polynomials

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