Differentiation of Different Type of Functions

Differentiation of Different Types of Functions: We use the concept of derivatives to express the rate of change in any function, such as identity, trigonometric, inverse trigonometric, polynomial, quadratic, and rational functions. A derivative considers even small changes in the dependent variable with respect to small changes in the independent variable.

In other words, if there is a small change in the dependent variable with respect to the independent variable, that change will be considered. The method of finding the derivative is known as differentiation. It has been used across fields like Science, Engineering, and Physics.

Differentiation of Different Types of Functions

Let \(A\) and \(B\) be two non-empty sets and let there exist a rule or manner or correspondence \(f\) which associates to each element of \(A\), a unique element in \(B\). Then \(f\) is called a function or mapping from \(A\) to \(B\). It is denoted as

\(f: A \rightarrow B\) or \(A \stackrel{f}{\rightarrow} B\)

It means that \(f\) is a function from \(A\) to \(B\) or \(f\) maps \(A\) to \(B\).

Derivatives of a Function

If \(f\) is a real-valued function, then the function defined by \(\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h}\), is defined to be the derivative of \(f\) at \(x\), wherever the limit exists. It is denoted as \(f^{\prime}(x)\). Therefore, we can say that

\(f’\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h}\)

provided that the limit exists. This method is known as the derivative of a function from the first principle.

Here \(f^{\prime}(x)=\frac{d}{d x}(f(x))\)

If \(y=f(x)\), then it is denoted by \(\frac{d y}{d x}\), i.e. derivative of \(f(x)\).

Derivative of Identity Functions

What is an Identity Function?

The function \(f: A \rightarrow B\) defined by \(f(x)=x ; x \in A\); (where \(A\) is a subset of \(B\)) is called the identity function.

Note:

1. If \(B=A\), then the function is called identity function on set \(A\) and is denoted by \(I_{A}\).
2. Identity function \(I_{A}\) defined on \(A\) is a bijection.

Derivative of \(x\)

Since identity function is of the form \(f(x)=x\), then derivative of identity function is \(\frac{d(f(x))}{d x}=1 ; x \in A\)

Derivatives of Trigonometric Functions

The trigonometric ratios are \(\sin x, \cos x, \tan x, \sec x, \operatorname{cosec} x\), and \(\cot x\).

Differentiation of \(\sin x\)

Let \(f(x)=\sin x\). Then, by definition, we have

\(f^{\prime}(x)=\frac{d}{d x}(f(x))=\mathop {\lim }\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}=\mathop {\lim }\limits_{h \to 0} \frac{\sin (x+h)-\sin x}{h}\)

\(=\mathop {\lim }\limits_{h \to 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h} \quad\left[\because \sin x-\sin y=2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\right]\)

\( = \mathop {\lim }\limits_{h \to 0} \cos \left( {x + \frac{h}{2}} \right)\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}\) As \(\left.h \rightarrow 0, \frac{h}{2} \rightarrow 0\right]\)

\(=\cos x \times 1\) \(\left[\because \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}=1\right]\)

\(=\cos x\)

Hence, \(\frac{d}{d x}(\sin x)=\cos x\)

Differentiation of \(\cos x\)

Let \(f(x)=\cos x\). Then, by definition, we have

\(f^{\prime}(x)=\frac{d}{d x}(f(x))=\mathop {\lim }\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}=\mathop {\lim }\limits_{h \to 0} \frac{\cos (x+h)-\cos x}{h}\)

\(=\mathop {\lim }\limits_{h \to 0} \frac{-2 \sin \left(\frac{2 x+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h} \quad\left[\because \cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\right]\)

\( = \mathop {\lim }\limits_{h \to 0} – \sin \left( {x + \frac{h}{2}} \right)\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}\) [As \(h \to 0,\,\frac{h}{2} \to 0\)]

\(=-\sin x \times 1\) \(\left[\because \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}=1\right]\)

\(=-\sin x \times 1\)

\(=-\sin x\)

Hence, \(\frac{d}{d x}(\cos x)=-\sin x\)

Differentiation of \(\tan x\)

Let \(f(x)=\tan x\). Then, by definition, we have

\(f^{\prime}(x)=\frac{d}{d x}(f(x))=\mathop {\lim }\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}=\mathop {\lim }\limits_{h \to 0} \frac{\tan (x+h)-\tan x}{h}\)

\(=\mathop {\lim }\limits_{h \to 0} \frac{\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}}{h}\quad\left[\because \tan x=\frac{\sin x}{\cos x}\right]\)

\(=\mathop {\lim }\limits_{h \to 0} \frac{[\sin (x+h) \cos x-\sin x \cos (x+h)]}{h \cos (x+h) \cos x}\)

\(=\mathop {\lim }\limits_{h \to 0} \frac{\frac{[\sin (2 x+h)+\sin h]}{2}-\frac{[\sin (2 x+h)-\sin h]}{2}}{h \cos (x+h) \cos x}\)

\(=\mathop {\lim }\limits_{h \to 0} \frac{\sin (h)}{h \cos (x+h) \cos x}\)

\(=\mathop {\lim }\limits_{h \to 0} \frac{1}{\cos (x+h) \cos x} \times \mathop {\lim }\limits_{h \to 0} \frac{\sin (h)}{h} \quad\left[\right.\) As \(\left.h \rightarrow 0, \frac{h}{2} \rightarrow 0, \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}=1\right]\)

\(=1 \times \frac{1}{\cos x \times \cos x}\)

\(=\frac{1}{\cos ^{2} x}=\sec ^{2} x\) \(\left[\because \sec x=\frac{1}{\cos x}\right]\)

Hence, \(\frac{d}{d x}(\tan x)=\sec ^{2} x\)

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

Derivatives of all Trigonometric Functions

1. \(\frac{d}{d x}(\sin x)=\cos x\)
2. \(\frac{d}{d x}(\cos x)=-\sin x\)
3. \(\frac{d}{d x}(\tan x)=\sec ^{2} x\)
4. \(\frac{d}{d x}(\cot x)=-\operatorname{cosec}^{2} x\)
5. \(\frac{d}{d x}(\sec x)=\sec x \tan x\)
6. \(\frac{d}{d x}(\operatorname{cosec} x)=-\operatorname{cosec} x \cot x\)

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 an angle whose sine is \(x\).

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

Two important notes related to inverse trigonometric function may be required while solving the questions related to differentiation.

• \(\mathop {\lim }\limits_{x \to 0} \frac{\sin ^{-1} x}{x}=1\)
• \(\mathop {\lim }\limits_{x \to 0} \frac{\tan ^{-1} x}{x}=1\)

Derivative of \(\sin ^{-1} x\)

Let \(y=\sin ^{-1} x\), where \(x \in(-1,1)\)

\(\Rightarrow \quad x=\sin y\)\(……(i)\)

Let \(\delta y\) be a small increment in \(y\) and \(\delta x\) the corresponding increment in \(x\)

\(\therefore x+\delta x=\sin (y+\delta y)\)\(……(ii)\)

Subtracting \((i)\) from \((ii)\), we get

\(\delta x=\sin (y+\delta y)-\sin y=2 \cos \left(y+\frac{\delta y}{2}\right) \cdot \sin \frac{\delta y}{2}\)\(……(iii)\)

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 \(\delta y \rightarrow 0\), we get

\(\mathop {\lim }\limits_{\delta y \to 0} \frac{\delta x}{\delta y}=\mathop {\lim }\limits_{\delta y \to 0} \cos \left(y+\frac{\delta y}{2}\right) \cdot \mathop {\lim }\limits_{\delta y \to 0} \frac{\sin \frac{\delta y}{2}}{\frac{\delta y}{2}}\) \(\left[\text {As}\, \delta y \rightarrow 0, \frac{\delta y}{2} \rightarrow 0\right]\)

\(\therefore \quad \frac{d x}{d y}=\cos y \cdot 1=\cos y \quad\left[\because \mathop {\lim }\limits_{x \to 0} \frac{\sin ^{-1} x}{x}=1\right]\)

i. e., \(\frac{d y}{d x}=\frac{1}{\cos y}=\frac{1}{\pm \sqrt{1-\sin ^{2} y}}=\pm \frac{1}{\sqrt{1-x^{2}}}\)

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 \(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}\)

\(\therefore \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}, x \in(-1,1)\)

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

Derivatives of Polynomials

What is a Polynomial?

An expression of the form \(a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\cdots+a_{n} x^{n}\left(a_{n} \neq 0\right)\), where \(a_{0}, a_{1}, a_{2}, a_{3}, \ldots a_{n}\) are real constants is a polynomial.
Here, \(n\) is the degree of the polynomial. It is also known as the \(n^{t h}\) degree polynomial.

Differentiation of Polynomials

Theorem: If \(f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\cdots+a_{n} x^{n}\left(a_{n} \neq 0\right)\), where \(a_{0}, a_{1}, a_{2}, a_{3}, \ldots a_{n}\) are real constants, then \(f^{\prime}(x)=a_{1}+2 a_{2} x+3 a_{3} x^{2}+\cdots+n a_{n} x^{n-1}\), i.e., the derivative of a polynomial of degree \(n\) is a polynomial of degree \(n-1\).

Proof: Let \(f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\cdots+a_{n} x^{n}\left(a_{n} \neq 0\right)\)

Therefore, \(f^{\prime}(x)=\frac{d}{d x}\left(a_{0}\right)+a_{1} \frac{d}{d x}(x)+a_{2} \frac{d}{d x}\left(x^{2}\right)+a_{3} \frac{d}{d x}\left(x^{3}\right)+\cdots+a_{n} \frac{d}{d x}\left(x^{n}\right)\)

We know that, \(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\) and \(\frac{d}{d x}(c)=0\), where \(c\) is a constant.

\(\Rightarrow f^{\prime}(x)=0+a_{1}(1)+a_{2}(2 x)+a_{3}\left(3 x^{2}\right)+\cdots+a_{n}\left(n x^{n-1}\right)\)

\(=a_{1}+2 a_{2} x+3 a_{3} x^{2}+\cdots+n a_{n} x^{n-1}\)

Here the degree of the polynomial after differentiation is \(n-1\).

If \(f(x)\) and \(g(x)\) are two differentiable polynomials, then:

1. \(\frac{d}{d x}(c f(x))=c f^{\prime}(x)\) where \(c\) is a constant.
2. \(\frac{d}{d x}[f(x) \pm g(x)]=f^{\prime}(x) \pm g^{\prime}(x)\)
3. \(\frac{d}{d x}[f(x) \times g(x)]=f^{\prime}(x) g(x)+g^{\prime}(x) f(x)\)
4. \(\frac{d}{d x}[f(x) / g(x)]=\left[g(x) f^{\prime}(x)-f(x) g^{\prime}(x)\right] /(g(x))^{2}\), where \(g(x) \neq 0\)

Derivative of Quadratic Function

What is a Quadratic Function?

A polynomial function of second-degree is known as a quadratic function. \(\Rightarrow p(x)=a_{0}+a_{1} x+a_{2} x^{2}\), where \(a_{2} \neq 0\) is a quadratic function.

Differentiation of Quadratic Functions

We know that derivative of a polynomial function with degree \(n\) is a polynomial function with degree \(n-1\).

The degree of a quadratic function is \(2\).

Therefore, the degree of derivative of a quadratic function is \(1\).

Hence, the derivative of a quadratic function is a linear function of the form \(a_{1}+2 a_{2} x\).

Derivative of Rational Function

What is a Rational Function?

The function that can be written as the quotient or ratio of two polynomial functions, where the denominator is not \(0\), is said to be a rational function.

Differentiation of Rational Functions

If \(P(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\cdots+a_{n} x^{n} ; Q(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+\cdots\) \(+b_{n} x^{n} \neq 0\) be two polynomial functions, then \(\frac{P(x)}{Q(x)}\) will be a rational function of \(x\). Since rational functions are of the form \(\frac{P(x)}{Q(x)}\) which is nothing but the \(\frac{p}{q}\) form, we know that

\(\frac{d\left(\frac{P(x)}{Q(x)}\right)}{d x}=\frac{Q(x) \cdot P^{\prime}(x)-P(x) Q^{\prime}(x)}{Q^{2}(x)}\)

where \(Q(x) \neq 0\).

Solved Example

Q1. Find the derivative of the function \(\sin 3 x \cos 3 x\) with respect to \(x\).
Ans: Let \(f(x)=\sin 3 x \cos 3 x\)
\(f(x)=\frac{2}{2} \sin 3 x \cos 3 x\)
\(f(x)=\frac{\sin 6 x}{2}\) \([\because 2 \sin x \cos x=\sin 2 x]\)
\(\Rightarrow f^{\prime}(x)=\frac{\cos 6 x}{2} \times 6\)
\(\therefore f^{\prime}(x)=3 \cos 6 x\)

Q2. Find the derivative of \(f(x)=x^{8}+4 x^{7}-2 x^{5}+10 x-16\), and hence find \(f^{\prime}(\mathbf{0})\).
Ans: Given: \(f(x)=x^{8}+4 x^{7}-2 x^{5}+10 x-16\)
Therefore, \(f^{\prime}(x)=\frac{d}{d x}\left(x^{8}\right)+\frac{d}{d x}\left(4 x^{7}\right)-\frac{d}{d x}\left(2 x^{5}\right)+\frac{d}{d x}(10 x)-\frac{d}{d x}(16)\)
\(=8 x^{7}+28 x^{6}-10 x^{4}+10\)
Now, \(f^{\prime}(0)=8(0)^{7}+28(0)^{6}-10(0)^{4}+10\)
\(\therefore f^{\prime}(0)=10\)

Q3. If \(y=\sin ^{-1}\left(2 x^{2}-1\right)\) find \(\frac{d y}{d x}\).
Ans: The given is inverse trigonometric. Hence, we can use the substitution method to find the derivative.
Given: \(y=\sin ^{-1}\left(2 x^{2}-1\right)\)
Put \(x=\cos \theta\), so that \(\theta=\cos ^{-1} x\)
\(y=\sin ^{-1}\left(2 \cos ^{2} \theta-1\right)\)
\(y=\sin ^{-1}(\cos 2 \theta)=\frac{\pi}{2}-2 \theta=\frac{\pi}{2}-2 \cos ^{-1} x\)
\(\therefore \frac{d y}{d x}=0+\frac{2}{\sqrt{1-x^{2}}}\)
Hence, the derivative of \(y=\sin ^{-1}\left(2 x^{2}-1\right)\) is \(+\frac{2}{\sqrt{1-x^{2}}}\).

Q4. Differentiate \(\frac{3+4 x}{2-x}\) w.r.t. \(x\)
Ans: Let \(y=\frac{3+4 x}{2-x}\)
Given function is a rational function.
Therefore, by using the quotient rule, we get
\(\frac{d y}{d x}=\frac{\left(\frac{d(3+4 x)}{d x} \cdot(2-x)-(3+4 x) \cdot \frac{d(2-x)}{d x}\right)}{(2-x)^{2}}\)
\(\frac{d y}{d x}=\frac{4(2-x)-(3+4 x)(-1)}{(2-x)^{2}}\)
\(\therefore \frac{d y}{d x}=\frac{11}{(2-x)^{2}}\)

Q5. Find the derivative of \(9 x^{2}-5 x+11\) with respect to \(x\) and hence find its value at \(x=-1\)
Ans: Let \(f(x)=9 x^{2}-5 x+11\)
Differentiating both sides w.r.t. \(x\), we get
\(f^{\prime}(x)=18 x-5\)
Put \(x=-1\)
We get, \(f^{\prime}(-1)=18(-1)-5=-23\)

Summary

The function \(f: A \rightarrow B\) defined by \(f(x)=x ; x \in A\); (where \(A\) is a subset of \(B\)) is called the identity function. The derivative of identity function is \(\frac{d(f(x))}{d x}=1 ; x \in A\).
Derivatives of all trigonometric functions are given as \(\frac{d}{d x}(\sin x)=\cos x, \frac{d}{d x}(\cos x)=\,-\sin x, \frac{d}{d x}(\tan x)=\sec ^{2} x, \frac{d}{d x}(\cot x)=-\operatorname{cosec}^{2} x,\)
\(\frac{d}{d x}(\sec x)=\sec x \tan x\), \(\frac{d}{d x}(\operatorname{cosec} x)=-\operatorname{cosec} x \cot x\).

For a polynomial \(f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\cdots+a_{n} x^{n}\left(a_{n} \neq 0\right)\), where \(a_{0}, a_{1}, a_{2}, a_{3}, \ldots a_{n}\) are real constants, the derivative is \(f^{\prime}(x)=a_{1}+2 a_{2} x+3 a_{3} x^{2}+\cdots+n a_{n} x^{n-1}\). The derivative of a polynomial of degree \(n\) is a polynomial of degree \(n-1\).

The degree of derivative of a quadratic function is \(1\). Hence, the derivative of a quadratic function is a linear function of the form \(a_{1}+2 a_{2} x\). Rational functions are of the form \(\frac{P(x)}{Q(x)}\) and \(\frac{d\left(\frac{P(x)}{Q(x)}\right)}{d x}=\frac{Q(x) \cdot P^{\prime}(x)-P(x) Q^{\prime}(x)}{Q^{2}(x)}\), where \(Q(x) \neq 0\).

Frequently Asked Questions (FAQs)

Below are the frequently asked questions on Differentiation of Different Type of Functions:

Q.1. How can one determine if an equation is a function or not?
Ans: For any function \(f: X \rightarrow Y\) to be a function, three conditions need to be satisfied are:
1. \(f \subset X \times Y\)
2. For each \(x \in X \Rightarrow(x, f(x)) \in f\)
3. \(\left(x, y_{1}\right) \in f\) and \(\left(x, y_{2}\right) \in f \Rightarrow y_{1}=y_{2}\)

Q.3. How can one determine if a function is a function?
Ans: There are different methods to check if a given equation is a function or not.
1. Examine the Order Pairs:
Check each ordered pair of \(y\) corresponding to \(x\) should be unique.
2. Solving for \(y\):
Corresponding to each value of \(x\), there should be only one value of \(y\).
3. Vertical Line Test:
A drawn vertical line must cross the graph of the relation exactly once.
4. Input-Output Chart:
Any input-output chart where thr input has two or more outputs is not a function.

Q.4. How do you differentiate a function?
Ans: First, identify the type of the given function. Then, by using suitable methods, differentiate the function.

Q.5. What is your opinion about a parent function?
Ans: The parent function is the simplest function of the family of functions that preserves the definition and properties of the family. With the help of a parent function, we can generate any function of the family by translation.
For example, the family of quadratic functions having the general form \(y=a x^{2}+b x+c, a \neq 0. y=x^{2}\) is the parent function.

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