Product Rule Formula: Derivation & Differentiation

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Jasmine Grover

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Product rule formula enables us to distinguish between two or more functions within a given function. The formula for the product rule is as follows: 

\(\frac{d(uv)}{dx} = u\frac{dv}{dx} + v \frac{du}{dx}\)

  • The product rule is revealed when multiplying the first function by the derivatives of the second plus the second function multiplied by the derivatives of the first function. 
  • Here, we assume that the first term of u constant and the second term of v constant are constants.
  • The Product Rule, or the Leibniz Rule, can be applied to determine the derivative of a given function of the form, f(x).g(x), in order for both f(x) and g(x) to become differentiable.
  • It follows the very concept seen in limits and derivatives.

Key Terms: Differentiation, Derivative, Limits, Product Rule, Leibniz Rule, Chain Rule, Lagrange’s Notation


What is Product Rule?

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"Product rule" in mathematics is used to determine the differential of any function presented in the form of a product formed by multiplying any two differentiable functions. 

  • The derivation of a product of two differentiable functions is equal to the sum of the first product function with the second function differentiation and the first product function with the second function differentiation, as per product rule.
  • This indicates that if a function of the format f(x).g(x) is given, we can use the product rule derivative to calculate the derivation of this function as, d / dx f(x)·g(x) = [g(x) × f'(x) + f(x) × g'(x)]

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Continuity and Differentiability Detailed Video Explanation:

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Product Rule Formula

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The Product Rule Formula can be expressed as:

\(\frac{d(uv)}{dx} = u\frac{dv}{dx} + v \frac{du}{dx}\)

Using the mathematics product rule formula, we can calculate the derivatives or assess the differentiation of the product of two functions. The following is the product rule formula of two functions:

d/dx f(x) = d / dx {u(x) · v(x)} = [v(x) × u'(x) + u(x) × v'(x)]

Here,

  • f(x) stands for Product of differentiable functions u(x) and v(x)
  • u(x), v(x) stands for Differentiable functions
  • u'(x) stands for Derivative of function u(x)
  • v'(x) stands for Derivative of the function v(x)

Derivation of Product Rule Formula

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The product rule can be expressed using Lagrange notation for any two functions as

In Leibniz's notation, this is written as (u v)' = u'·v + u·v'.

d / dx (u·v) = du / dx ·v + u·dv / dx

Let's examine this evidence for the product rule formula. There are various ways to demonstrate the product rule formula, including:

  • Using the First principle
  • With the Chain rule

Product Rule Formula using First Principle

Let the function h(x) = f(x)g(x), such that f(x) and g(x) are differentiable at x, be used to demonstrate the product rule formula using the definition of derivative or limits.

 h'(x) = limΔx→0 [h(x + Δx) - h(x)]/Δx

= limΔx→0 f(x+Δx)g(x+Δx)−f(x)g(x) / Δx

= limΔx→0 f(x+Δx)g(x+Δx)−f(x)g(x+Δx)+f(x)g(x+Δx)−f(x)g(x) / Δx

= limΔx→0 [f(x+Δx)−f(x)]g(x+Δx)+f(x)[g(x+Δx)−g(x)] / Δx

= limΔx→0 [f(x+Δx)−f(x)]g(x+Δx) / Δx + limΔx→0 f(x)[g(x+Δx)−g(x)] / Δx 

= (limΔx→0 [f(x+Δx)−f(x)] / Δx) (limΔx→0 g(x+Δx))+(limΔx→0 f(x)) (limΔx→C

= g(x)limΔx→0 [f(x+Δx)−f(x)] / Δx + f(x) lim Δx→0 [g(x+Δx)−g(x)] / Δx 

∵ limΔx→0 [f(x+Δx)−f(x)] / Δx limΔx→0 [g(x+Δx)−g(x)] / Δx = g'(x)

= d / dx f(x)·g(x) = [g(x) × f'(x) + f(x) × g'(x)]

Hence, proved.

Product Rule Formula using Chain Rule

By thinking of the product rule as a specific case of the chain rule, you can use the chain rule formula to obtain the mathematics formula for the product rule. If f(x) is a differentiable function, then h(x) must equal f(x)g (x).

d / dx (f·g) = [δ(fg) / δf][df / dx] + [δ(fg) / δg][dg / dx] = g(df / dx) + f(dg / dx)

Product rule for Product of more than Two Functions

Through the same technique, the product rule can be expanded to encompass products of more than two components. For instance, assuming that the product of three functions, u(x), v(x), and w(x), is considering as u(x)v(x)w(x), we obtain,

d(uv w) / dx = vw du/dx + u dv/dx w + uv dw/dx d(uvw) / dx = du/dx vw + u d/d

Frequently Asked Question

Ques: When is Product Rule Used?

Ans: When working with the products of two functions, the product rule is used. While the chain rule is used when distinguishing a composite function, also referred to as a function of a function, such as f(x).


Product Rule in Differentiation

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In order to determine the derivative of the function h(x) = f(x)g(x), both f(x) and g(x) have to be differentiable functions. By using the product rule, we can apply the following steps to find the derivation of a differentiable function h(x) = f(x)g(x):

Step 1: Check the values of f(x) and g(x).

Step 2: Now, determine the values of f'(x) and g'(x) and further apply the product rule formula, which can be expressed as: h'(x) = \(\frac{d}{dx}\)f(x)·g(x) = [g(x) × f'(x) + f(x) × g'(x)]

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Things to Remember

  1. A mathematics technique known as the "product rule" can be used to determine the derivative or differentiation of a function that is given as the combination of two differentiable functions.
  2. The product rule effectively adopts the differentiations notions of limits and derivatives.
  3. The product rule of differentiation or (dy/dx) = u (dv/dx) + v (du/dx) is also known as the product rule for derivatives.
  4. Gottfried Leibniz's product rule enables us to compute derivatives that we don't want (or are unable to) multiply swiftly.

Previous Year Questions


Sample Questions 

Ques 1. Describe the primary principle of differentiation. (1 mark)

Ans. At every point, we can define the derivative of function F. If the derivative is present everywhere, it becomes a new function represented by the symbol f'. Remember that perhaps the area where the function f'(x) exists is the domain of the derivative. It's possible to shorten f '(x) to d/dx (f(x)) or dy / dx.

Ques 2. Differentiate the following function: ( x2 + 3)(5x + 4) (3 marks)

Ans. The function that is given is (x2 + 3)(5x + 4)

Here, u equals (x2 + 3) and v equals (5x + 4)

Applying the product rule

D((X2 + 3)(5x + 4)) / dx

= (x2 + 3) d(5x + 4) / dx + (5x + 4) d(x2 + 3) / dx

= (x2 + 3)5 + (5x + 4) 2x

= 5x2 + 15 + 10x2 + 8x

= 15x2 + 8x + 15

Ques 3. What do you mean by product rule maths? (2 marks)

Ans. Only when the two "components" of the function are multiplied together is the product rule applied, and when they are being combined, the chain rule is applied. 

For instance, we can use the product rule to determine the derivative of f(x) = x2 sin(x) and the chain rule to determine the derivative of g(x) = sin(x2).

Ques 4. Utilising the product rule, locate f'(x) for the following function f(x) = xlog x. (3 marks)

Ans. Given in the question,

f(x) = x·log x

u(x) = x
v(x) = log x

⇒u'(x) = 1
⇒v'(x) = 1/x

⇒f'(x) = [v(x)u'(x) + u(x)v'(x)]
⇒f'(x) = [log x•1 + x•(1/x)]
⇒f'(x) = log x + 1

Ques 5. Analyze the function y = x(1 + x) to find its derivative. (2 marks)

Ans. By using product rule,

y'(x) = {x(1 + x)}’ = x'(1 + x) + x(1 + x)’ 

= (1 + x) + x(0 + 1) 

= 1 + 2x

Ques 6. Assume y = cos2x. Make this function distinct by applying the product rule. (2 marks)

Ans. Given in the question,

y(x) = cosxcosx .

By using product rule,

y′(x)= (cosx cosx)′ = (cosx)′cosx + cosx(cosx)′.

Since (cosx)′ = -sinx, we obtain

y′(x) = – sinxcosx + cosx(-sinx) = – 2sinxcosx = – sin2x

Ques 7. Discover the derivative of the equation y = exsinx. (2 marks)

Ans. Here,

y′(x) = (exsinx)′ = (ex)′sinx + ex(sinx)′ 

= exsinx + ex(cosx)

= ex(sinx + cosx).

Ques 8. Analyze the function y = xsinx's derivative. (2 marks)

Ans. By using the product rule we get,

y′(x) = (x sinx)′ = (x)′sinx + x (sinx)′ 

= sinx + x cosx 

Ques 9. What is the derivative of x· cos(x)? Use the product rule formula to determine the same. (3 marks)

Ans.Let f(x) = cos x 

And, let g(x) = x.

Thus, 

⇒f'(x) = -sin x

⇒g'(x) = 1

⇒[f(x)g(x)]' = [g(x)f'(x) + f(x)g'(x)]

⇒[f(x)g(x)]' = [(x•(-sin x) + cos x•(1)]

Hence,

⇒[f(x)g(x)]' = - x sin x + cos x

Ques 10. Apply the product rule formula in order to differentiate (1 - 2x)·sin x. (3 marks)

Ans. Let f(x) = (1-2x), 

And Let g(x) = sin x

Thus,

⇒ g'(x) = cos x

⇒ f'(x) = -2

⇒ [f(x)g(x)]' = [g(x)f'(x) + f(x)g'(x)]

⇒ [f(x)g(x)]' = [(sin x•(-2) + (1 - 2x)•(cos x)]

Hence,

⇒ [f(x)g(x)]' = - 2 sin x + cos x - 2x cos x

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CBSE CLASS XII Related Questions

  • 1.
    Find:

    If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

      • \(0\)
      • \(-2\)
      • \(-1\)
      • \(2\)

    • 2.
      If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).


        • 3.
          Find:

          The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]


            • 4.

              Evaluate:
              \[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]


                • 5.
                  Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]


                    • 6.
                      Find:

                      If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

                        • \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
                        • \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
                        • \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
                        • \(p = 0, \, q = 0\)
                      CBSE CLASS XII Previous Year Papers

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