Algebra of Derivatives of Functions: The theory of the concept of derivatives is a bit abstract. The derivative is the fundamental and very important topic of Calculus whereas, the derivative of functions follow the concept of limits. It has numerous applications in real-life scenarios, including calculating rates of change. For instance, the rate of change of displacement of a spaceship is its velocity, and the rate of change of its velocity is its acceleration required to reach a particular point in a given time.

With the basic knowledge of the algebra of functions, along with the three basic derivatives and the four rules of algebraic operations of derivatives, one should be able to get a good grip of the topic. The basic rules explained in this article can act as a foundation to build advanced studies related to derivatives.

Definition of Derivatives

The derivative of a real-valued function measures the tendency of the function to change values with respect to the change in its independent variable.

If \(f\left( x \right)\) is a real-valued function that is differentiable at a point \(a,\) and if the domain contains an open interval \(I\) containing \(a,\) and the limit \(L = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({a + h} \right) – f\left( a \right)}}{h}\) exists, this limit is called the derivative of \(f\left( x \right)\) at \(a.\)
The function \(f\left( x \right)\) is said to be differentiable in the interval \(I,\) if it is differentiable at every point of \(I.\) The derivative of the function \(f\left( x \right)\) is denoted as:
\(\frac{d}{{dx}}\left[{f\left( x \right)} \right]\) or \(f’\left( x \right).\)

Derivatives as Slope of the Tangent Line

The slope of a tangent line at a point on the function is equal to the derivative of the function at that point.

Here, the derivative of the function \(f\left( x \right)\) at the point \({x_1}\) or \(f’\left({{x_1}} \right)\) is the slope of the tangent line at the point \({x_1}.\) Similarly, the derivatives at the points \({x_2}\) and \({x_3}\) are the slopes of the tangent lines at the respective points, as shown in the figure.

Let us understand the basic algebraic operations and rules related to derivatives of functions.

Algebra of Derivatives of Functions: Important Rules

1. Sum Rule

The derivative of the sum of two functions can be calculated as the sum of their derivatives. That is, if \(f\left( x \right)\) and \(g\left( x \right)\) are two real-valued differentiable functions in the interval \(I,\) then,
\(\frac{d}{{dx}}\left[{f\left( x \right) + g\left(x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x \right)} \right] + \frac{d}{{dx}}\left[{g\left(x \right)} \right]\)

Proof: The sum rule can be proved using the properties of limits and the definition of derivatives as \(f’\left( x \right) = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right) – f\left( x \right)}}{h}\)
\(\therefore \frac{d}{{dx}} \left[{f\left( x \right) + g\left( x \right)} \right]=\underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right) + g\left({x + h} \right) – \left[{f\left( x \right) + g\left( x \right)} \right]}}{h}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right) – f\left( x \right) + g\left({x + h} \right) – g\left( x \right)}}{h}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h}\right) – f\left( x \right)}}{h} + \underset{{h \to 0}}{\mathop {\lim }} \frac{{g\left({x + h} \right) – g\left( x \right)}}{h}\)
\( = \frac{d}{{dx}}\left[{f\left( x \right)} \right] + \frac{d}{{dx}}\left[{g\left( x \right)} \right]\)
Thus, we have, \(\frac{d}{{dx}}\left[{f\left( x \right) + g\left( x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x \right)} \right] + \frac{d}{{dx}}\left[{g\left( x \right)} \right]\)

2. Difference Rule

Similar to the sum rule, the derivative of the difference of two functions can also be calculated as the difference of their derivatives. That is, if \(f\left( x \right)\) and \(g\left( x \right)\) are two real-valued differentiable functions in the interval \(I,\) then,
\(\frac{d}{{dx}}\left[{f\left( x \right) – g\left(x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x \right)} \right] – \frac{d}{{dx}}\left[{g\left(x \right)} \right]\)
Proof: The difference rule can be proved using the properties of limits and the definition of derivatives as \(f’\left( x \right) = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right) – f\left( x \right)}}{h}\)
\(\therefore \frac{d}{{dx}} \left[{f\left( x \right)} \right] – \frac{d}{{dx}}\left[{g\left( x \right)} \right]=\underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right) – f\left( x \right)}}{h}\underset{{h \to 0}}{\mathop {\lim }} \frac{{g\left({x + h} \right) – g\left(x \right)}}{h}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h}\right) – f\left( x \right) – g\left({x + h}\right) + g\left( x \right)}}{h}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left( {x + h} \right) – g\left( {x + h} \right) – \left[ {f\left( x \right) – g\left( x \right)} \right]}}{h}\)
\(=\frac{d}{{dx}}\left[ {f\left( x \right) – g\left( x \right)} \right]\)
Hence, we have, \(\frac{d}{{dx}}\left[{f\left( x \right) – g\left( x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x \right)} \right] – \frac{d}{{dx}}\left[{g\left( x \right)} \right]\)

3. Product Rule

The product rule states that the derivative of the product of two functions is the sum of the derivative of the first function times the second function, and the first function times the derivative of the second function.
That is, if \(f\left( x \right)\) and \(g\left( x \right)\) are two real-valued differentiable functions in the interval \(I\) then,

\(\frac{d}{{dx}}\left[{f\left( x \right).g\left( x \right)} \right] = \frac{d}{{dx}}\left[ {f\left( x \right)} \right].g\left( x \right) + f\left( x \right).\left[{\frac{d}{{dx}}g\left( x \right)} \right]\)

Proof:
The product rule can also be proved using the properties of limits and the definition of derivatives as \(f’\left(x \right) = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right) – f\left( x \right)}}{h}.\)
\(\therefore\frac{d}{{dx}}\left[{f\left(x \right).g\left( x \right)}\right] = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right)g\left({x + h} \right) – f\left( x \right)g\left( x \right)}}{h}\)
Adding and subtracting \(f\left({x + h} \right)g\left( x\right)\) in the numerator, we get,
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right)g\left({x + h} \right) – f\left({x + h} \right)g\left( x \right) + f\left({x + h} \right)g\left( x \right) – f\left( x \right)g\left( x \right)}}{h}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right)\left[{g\left({x + h} \right) – g\left( x \right)} \right] + g\left(x \right)\left[{f\left({x + h}\right) – f\left( x \right)} \right]}}{h}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} f\left({x + h} \right)\frac{{\left[{g\left({x + h} \right) – g\left( x \right)} \right]}}{h} + \underset{{h \to 0}}{\mathop {\lim }} g\left( x \right)\frac{{\left[{f\left({x + h} \right) – f\left( x \right)} \right]}}{h}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} f\left({x + h} \right)\underset{{h \to 0}}{\mathop {\lim }} \frac{{\left[{g\left({x + h}\right) – g\left(x \right)} \right]}}{h} + \underset{{h \to 0}}{\mathop {\lim }} g\left( x \right)\underset{{h \to 0}}{\mathop {\lim }} \frac{{\left[{f\left({x + h} \right) – f\left( x \right)} \right]}}{h}…..\left({\text{i}} \right)\)
Here, using the definition, we can say that,
\(\underset{{h \to 0}}{\mathop {\lim }} f\left({x + h} \right) = f\left( x \right),\,\,\,\underset{{h \to 0}}{\mathop {\lim }} \frac{{\left[{g\left({x + h} \right) – g\left( x \right)} \right]}}{h} = \frac{d}{{dx}}g\left( x \right)\)
\(\underset{{h \to 0}}{\mathop {\lim }} g\left( x \right)\,=\,g\left( x \right),\,\,\,\underset{{h \to 0}}{\mathop {\lim }} \frac{{\left[{f\left({x + h} \right) – f\left( x \right)} \right]}}{h} = \frac{d}{{dx}}f\left( x\right)\)
Applying these in \(\left({\text{i}} \right),\)
\( = f\left( x \right) \cdot \left[ {\frac{d}{{dx}}g\left( x \right)} \right] + \frac{d}{{dx}}\left[ {f\left( x \right)} \right] \cdot g\left( x \right)\)
Therefore, we have, \( \frac{d}{{dx}}\left[{f\left( x \right).g\left( x \right)}\right] = \frac{d}{{dx}}\left[{f\left( x \right)} \right].g\left( x \right) + f\left( x \right).\left[{\frac{d}{{dx}}g\left( x \right)} \right]\)
This rule is also known as the Leibnitz Rule of differentiation.

4. Quotient Rule

When two differentiable functions divide each other, the derivative is defined by the quotient rule of differentiation. The rule states that, if \(f\left( x \right)\) and \(g\left( x \right)\) are two real-valued differentiable functions in the interval \(I,\) then,

\(\frac{d}{{dx}}\left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{g\left( x \right)\frac{d}{{dx}}\left[ {f\left( x \right)} \right] – f\left( x \right)\frac{d}{{dx}}\left[ {g\left( x \right)} \right]}}{{{{\left[ {g\left( x \right)} \right]}^2}}}\)

Proof:
The quotient rule is also be derived using the properties of limits and the definition of derivatives as \(f’\left( x \right) = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({x + h} \right) – f\left( x \right)}}{h}.\)
\(\therefore \frac{d}{{dx}}\left[{\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \underset{{h \to 0}}{\mathop {\lim }} \frac{{\frac{{f\left({x + h}\right)}}{{g\left({x + h} \right)}} – \frac{{f\left(x \right)}}{{g\left(x \right)}}}}{h}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{1}{h}\,\frac{{f\left({x + h} \right)g\left( x \right) – f\left( x \right)g\left({x + h} \right)}}{{g\left( x \right)g\left({x + h} \right)}}\)
Subtracting and adding \(f\left( x \right)g\left( x \right)\) in the numerator,
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{1}{h}\,\frac{{f\left({x + h} \right)g\left( x \right) + f\left( x \right)g\left( x \right) – f\left( x \right)g\left({x + h} \right)}}{{g\left( x \right)g\left({x + h} \right)}}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{1}{{g\left( x \right)g\left({x + h} \right)}}\frac{{f\left({x + h} \right)g\left( x \right) – f\left( x \right)g\left( x \right) + f\left( x \right)g\left( x \right) – f\left( x \right)g\left({x + h}\right)}}{h}\)
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{1}{{g\left( x \right)g\left({x + h} \right)}}\left[{\frac{{f\left({x + h} \right)g\left( x\right) – f\left( x \right)g\left(x \right)}}{h} + \frac{{f\left(x \right)g\left( x\right) – f\left( x \right)g\left({x + h} \right)}}{h}} \right]\)
\( = \underset{{h \to 0}}{\mathop {\lim }} \frac{1}{{g\left( x \right)g\left({x + h} \right)}}\left[{g\left( x \right)\frac{{f\left({x + h} \right) – f\left( x \right)}}{h} – f\left( x \right)\frac{{g\left({x + h} \right) – g\left( x \right)}}{h}} \right]\)
By the basic properties of limits, we have,
\( = \frac{1}{{\underset{{h \to 0}}{\mathop {\lim }} g\left({x + h} \right)\underset{{h \to 0}}{\mathop {\lim }} g\left( x \right)}}\left[{\underset{{h \to 0}}{\mathop {\lim }} g\left(x \right)\underset{{h \to 0}}{\mathop{\lim }} \frac{{f\left({x + h} \right) – f\left( x \right)}}{h} – \underset{{h \to 0}}{\mathop {\lim }} f\left( x \right)\underset{{h \to 0}}{\mathop {\lim }} \frac{{g\left({x + h} \right) – g\left( x \right)}}{h}} \right]…\left({{\text{ii}}}\right)\)
Hence,
\( \underset{{h \to 0}}{\mathop {\lim }} f\left({x + h} \right) = f\left( x \right),\,\,\,\underset{{h \to 0}}{\mathop {\lim }} \frac{{\left[{f\left({x + h} \right) – f\left( x \right)} \right]}}{h} = \frac{d}{{dx}}f\left( x \right)\)
\( \underset{{h \to 0}}{\mathop {\lim }} g\left({x + h} \right) = g\left( x \right),\,\,\,\underset{{h \to 0}}{\mathop {\lim }} \frac{{\left[{g\left({x + h} \right) – g\left( x \right)} \right]}}{h} = \frac{d}{{dx}}g\left( x \right)\)
\( \underset{{h \to 0}}{\mathop {\lim }} f\left( x \right) = f\left( x \right),\,\,\,\underset{{h \to 0}}{\mathop {\lim }} g\left( x \right) = g\left( x \right)\)
Applying these in \(\left({{\text{ii}}} \right),\) We get,
\( = \frac{1}{{g\left( x \right).g\left( x\right)}}\left[{g\left( x \right).\frac{d}{{dx}}f\left( x \right) – f\left( x \right).\frac{d}{{dx}}g\left( x \right)} \right]\)
\( = \frac{{g\left( x \right)\frac{d}{{dx}}\left[ {f\left( x \right)} \right] – f\left( x \right)\frac{d}{{dx}}\left[ {g\left( x \right)} \right]}}{{{{\left[ {g\left( x \right)} \right]}^2}}}\)
Therefore, we get, \(\frac{d}{{dx}}\left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{g\left( x \right)\frac{d}{{dx}}\left[ {f\left( x \right)} \right] – f\left( x \right)\frac{d}{{dx}}\left[ {g\left( x \right)} \right]}}{{{{\left[ {g\left( x \right)} \right]}^2}}}\)

Other Important Rules Related to Derivatives

5. Constant Rule

Statement: The derivative of any constant is zero.
That is, \(\frac{d}{{dx}}\left( c \right) = 0\) for any constant \(c.\)
Moreover, the derivative of the product of any constant \(c\)
with \(x\) is \(c.\)
That is, \(\frac{d}{{dx}}\left( {cx} \right) = c\frac{d}{{dx}}\left( x \right) = c.\)

6. Power Rule

Statement: The derivative of \({x^n}\) is \(n{x^{n – 1}}.\)
That is, \(\frac{d}{{dx}}\left({{x^n}} \right) = n{x^{n – 1}}.\)
Further, this rule can be extended to form the general power rule that talks about the derivative of the \({n^{th}}\) power of a function.
The derivative of the \({n^{th}}\) power of a function is defined as:
\(\frac{d}{{dx}}\left[{{{\left({f\left( x \right)} \right)}^n}} \right] = n{\left({f\left( x \right)} \right)^{n – 1}}.f’\left( x \right)\)

7. Chain Rule

The chain rule is used to find the derivative of composite functions. If a function \(h\) is a composite function of \(f\) and \(g,\) the composite function \(\left({f \circ g} \right)\left( x \right)\) is calculated for a value of \(x\) by first evaluating \(g\left( x \right)\) and then calculating the value of the function \(f\) at \(g\left( x \right).\) Now, the chain rule explains the method of calculating the derivative of \(\left({f \circ g} \right)\left( x \right)\) as the product of the derivative of the function \(f\) at \(g\left( x \right)\) times the derivative of \(g\left( x \right).\)
That is, \(\frac{d}{{dx}}\left[ {f \circ g\left( x \right)} \right] = \frac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] \cdot \frac{d}{{dx}}g\left( x \right)\)

These results, along with Rolle’s mean value theorem, are known as important \(8\) theorems of derivatives.

Solved Examples – Algebra of Derivative of Functions

Q.1. Find the derivative of \(f\left( x \right) = 4{x^2} + 3x – 1.\)
Ans: To find the derivative of an algebraic expression, apply the sum and difference rules of derivatives.
\(f’\left( x \right) = \frac{d}{{dx}}\left({f\left( x \right)} \right)\)
\( = \frac{d}{{dx}}\left({4{x^2} + 3x – 1} \right)\)
\( = \frac{d}{{dx}}\left({4{x^2}} \right) + \frac{d}{{dx}}\left({3x} \right) – \frac{d}{{dx}}\left( 1 \right)\)
Applying the constant rule of derivatives,
\( = 4\frac{d}{{dx}}\left({{x^2}} \right) + 3\frac{d}{{dx}}\left( x \right) – 0\)
Applying the power rule of derivatives,
\( = 4\left({2x} \right) + 3\left( 1 \right)\)
\( = 8x + 3\)
Hence, \(f’\left( x \right) = 8x + 3.\)

Q.2. Prove that the derivative of \(g\left( x \right) = x\,\cos \,x\) is \(\cos x – x\,\sin \,x.\)
Ans: The given function is a product of two functions \(x\) and \(\cos \,x.\)
Hence, apply the product rule.
\(g’\left( x \right) = \frac{d}{{dx}}\left({g\left( x \right)} \right)\)
\( = \frac{d}{{dx}}\left({x\,\cos \,x} \right)\)
\( = x.\frac{d}{{dx}}\left({\cos \,x} \right) + \cos \,x\frac{d}{{dx}}\left( x \right)\)
\( = x.\left({ – \sin \,x} \right) + \cos \,x.1\)
\( = – x\,\sin \,x + \cos \,x\)
\(\therefore g’\left( x \right) = \cos \,x – x\,\sin \,x\)
Hence, proved.

Q.3. What is the derivative of the function \(k\left( x \right) = {\sin ^2}\,x\)?
Ans: The given function is a power of a base function. So, apply the general power rule that states:
\(\frac{d}{{dx}}{\left({k\left( x\right)} \right)^n} = n{\left[{k\left( x \right)} \right]^{n – 1}}k’\left( x \right)\)
Here,
\(k\left( x \right) = \sin \,x\)
\(n = 2\)
\(\therefore \frac{d}{{dx}}\left({{{\sin }^2}x} \right) = 2{\left({\sin \,x} \right)^{2 – 1}}.\frac{d}{{dx}}\left({\sin \,x} \right)\)
\( = 2\,\sin \,x\,\cos \,x\)
\( = \,\sin \,2x\)
\(\therefore k’\left( x \right) = \sin \,2x.\)

Q.4. Find \(h’\left( x \right),\) if \(h\left( x \right) = \frac{{\tan \,x}}{{{x^2}.}}\)
Ans: The function is given as a quotient of two functions. So, apply the quotient rule:
\(\frac{d}{{dx}}\left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{f’\left( x \right)g\left( x \right) – g’\left( x \right)f\left( x \right)}}{{g{{\left( x \right)}^2}}}\)
Here,
\(f\left( x\right) = \tan \,x\)
\(g\left( x \right) = \,{x^2}\)
\(\therefore h’\left( x \right) = \frac{{\left({\tan \,x} \right)’\left({{x^2}} \right) – \left({{x^2}}\right)’\left({\tan \,x} \right)}}{{{{\left({{x^2}} \right)}^2}}}\)
\( = \frac{{{x^2}{{\sec }^2}x – 2x\,\tan \,x}}{{{x^4}}}\)
\(\therefore h’\left( x \right) = \frac{{x\,{{\sec }^2}x – 2\,\tan \,x}}{{{x^3}}}\)

Q.5. What is the derivative of a function that is defined as its reciprocal?
Ans: The reciprocal function is defined as \(f\left( x \right) = \frac{1}{x}.\)
Rewriting the function as a power, \(\frac{1}{x} = {x^{ – 1}}.\)
Applying the power rule, we get,
\(f’\left( x \right) = \, – 1{x^{ – 1 – 1}}\)
\( = \,- 1{x^{ – 2}}\)
\( = \,- \frac{1}{{{x^2}}}\)

Summary

The article talks about the importance of knowing how to handle derivatives of functions that have many practical applications. It then elaborates the basic definition of the derivative of a function. Furthermore, the article moves on to the algebra of derivatives that explains the derivatives of real-valued differentiable functions with basic operations – addition, subtraction, multiplication, and division.

It elaborates the derivation of the four rules – sum rule, difference rule, product rule, and quotient rule. It also goes on to list a few other important rules such as – constant rule, power rule, general power rule, and chain rule. Then, it concludes with a few examples to understand the calculations involved in the algebra of derivatives of functions.

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Frequently Asked Questions (FAQs)

Q.1.How do you find the derivative of a function in a function?
Ans: The derivative of the function of a function can be calculated using the chain rule of differentiation, which states that:
\(\frac{d}{{dx}}\left[{f \circ g\left( x \right)} \right] = \frac{d}{{dx}}\left[{f\left({g\left( x \right)} \right)} \right].\frac{d}{{dx}}g\left( x \right)\)
For example, find the derivative of \(f \circ g\left( x \right),\) if \(f\left( x \right) = 5x – 3\) and \(g\left( x \right) = {x^2}.\)
\(f’\left( x \right) = 5\) and \(g’\left( x \right) = 2x\)
Then, the function \(f \circ g\left( x \right) = 5{x^2} – 3.\)
\(\frac{d}{{dx}}\left[{f \circ g\left( x \right)} \right] = f’\left({{x^2}} \right).2x\)
\( = 5\left({2x} \right)\)
\( = 10\,x\)

Q.2. What is the algebra of differentiation?
Ans: The basic operations on derivatives is also known as algebra of differentiation.
1. Sum Rule: \(\frac{d}{{dx}}\left[{f\left( x \right) + g\left(x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x\right)} \right] + \frac{d}{{dx}}\left[{g\left( x\right)} \right]\)
2. Difference Rule: \(\frac{d}{{dx}}\left[{f\left( x \right) – g\left(x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x\right)} \right] – \frac{d}{{dx}}\left[{g\left( x\right)} \right]\)
3. Product Rule: \(\frac{d}{{dx}}\left[{f\left( x \right).g\left( x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x \right)} \right].g\left( x \right) + f\left( x \right).\left[{\frac{d}{{dx}}g\left( x \right)} \right]\)
4. Quotient Rule: \(\frac{d}{{dx}}\left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{\frac{d}{{dx}}\left[{f\left( x \right)} \right]g\left( x \right) – f\left( x \right)\frac{d}{{dx}}\left[{g\left( x \right)} \right]}}{{g\left({{x^2}} \right)}}\)

Q.3. What is a derivative of a function in calculus?
Ans: If \(f\left( x \right)\) is a real-valued function that is differentiable at a point a and if the domain contains an open interval \(I\) containing a and the limit \(L = \underset{{h \to 0}}{\mathop {\lim }} \frac{{f\left({a + h} \right) – f\left( a \right)}}{h}\) exists, this limit is called the derivative of the function \(f\left( x \right)\) at \(a.\)

Q.4. What are the 7 differentiation rules?
Ans: The basic seven rules of differentiation are:
i. Sum Rule:
\(\frac{d}{{dx}}\left[{f\left( x \right) + g\left( x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x \right)}\right] + \frac{d}{{dx}}\left[{g\left( x \right)} \right]\)
ii. Difference Rule:
\(\frac{d}{{dx}}\left[{f\left( x \right) – g\left( x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x \right)}\right] – \frac{d}{{dx}}\left[{g\left( x \right)} \right]\)
iii. Product Rule:
\(\frac{d}{{dx}}\left[{f\left( x \right).g\left( x \right)} \right] = \frac{d}{{dx}}\left[{f\left( x \right)}\right].g\left( x \right) + f\left( x \right).\left[{\frac{d}{{dx}}g\left( x \right)} \right]\)
iv. Quotient Rule:
\(\frac{d}{{dx}}\left[{\frac{{f\left( x \right)}}{{g\left( x \right)}}}\right] = \frac{{\frac{d}{{dx}}\left[{f\left( x \right)} \right]g\left( x \right) – f\left( x \right)\frac{d}{{dx}}\left[{g\left( x \right)} \right]}}{{g{{\left( x \right)}^2}}}\)
v. Constant Rule: \(\frac{d}{{dx}}\left( c \right) = 0\)
vi. Power Rule: \(\frac{d}{{dx}}\left({{x^n}} \right) = n{x^{n – 1}}\)
vii. Chain Rule: \(\frac{d}{{dx}}\left[{f \circ g\left( x \right)}\right] = \frac{d}{{dx}}\left[{f\left({g\left( x \right)} \right)} \right].\frac{d}{{dx}}g\left( x \right)\)

Q.5. How does the product rule work?
Ans: When there is a product of two functions whose derivative needs to be calculated, then the product rule of derivatives is to be used. According to the product rule, the derivative of a product two functions is the sum of the derivative of the first function times the second and the first function times the derivative of the second function.
For example, the derivative of the function \(f\left( x \right) = \left({3{x^2} + 5x – 1} \right)\left({7x + 2}\right)\) can be calculated as:
\(f’\left( x \right) = \left({\frac{d}{{dx}}\left[{3{x^2} + 5x – 1} \right]} \right).\left({7x + 2} \right) + \left({3{x^2} + 5x – 1} \right).\left({\frac{d}{{dx}}\left[{7x – 2} \right]} \right)\)
\( = \left({6x + 5} \right).\left({7x + 2} \right) + \left({3{x^2} + 5x – 1} \right).\left( 7 \right)\)
\( = 4{x^2} + 12x + 35x + 10 + 21{x^2} + 35x – 7\)
\( = 63{x^2} + 82x + 3\)