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Q39:
Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ (2x – 7)^2 (3x + 5)^3 $

Solution :

Given Variables & Initial Setup

Let the given function be represented as $f(x)$:

$f(x) = (2x - 7)^2 (3x + 5)^3$

To differentiate this function with respect to $x$, we utilize the Product Rule [If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$] and the Chain Rule [If $y = g(h(x))$, then $\frac{dy}{dx} = g'(h(x)) \cdot h'(x)$].

Step 1: Identifying Components for the Product Rule

Let:

  • $u(x) = (2x - 7)^2$
  • $v(x) = (3x + 5)^3$

Step 2: Differentiating Components via Chain Rule

Differentiating $u(x)$:

$u'(x) = \frac{d}{dx}[(2x - 7)^2] = 2(2x - 7) \cdot \frac{d}{dx}(2x - 7)$ [By Power Rule and Chain Rule]

$u'(x) = 2(2x - 7) \cdot (2) = 4(2x - 7)$

Differentiating $v(x)$:

$v'(x) = \frac{d}{dx}[(3x + 5)^3] = 3(3x + 5)^2 \cdot \frac{d}{dx}(3x + 5)$ [By Power Rule and Chain Rule]

$v'(x) = 3(3x + 5)^2 \cdot (3) = 9(3x + 5)^2$

Step 3: Applying the Product Rule

$f'(x) = u'(x)v(x) + u(x)v'(x)$

$f'(x) = [4(2x - 7)] \cdot (3x + 5)^3 + (2x - 7)^2 \cdot [9(3x + 5)^2]$

Step 4: Algebraic Simplification and Factorization

Factor out the greatest common factors, which are $(2x - 7)$ and $(3x + 5)^2$:

$f'(x) = (2x - 7)(3x + 5)^2 [4(3x + 5) + 9(2x - 7)]$

Expand the terms inside the square brackets:

$f'(x) = (2x - 7)(3x + 5)^2 [12x + 20 + 18x - 63]$

Combine like terms:

$f'(x) = (2x - 7)(3x + 5)^2 [30x - 43]$

Input: f(x) = (2x-7)²(3x+5)³ Differentiation Output f'(x) = (2x-7)(3x+5)²(30x-43)

Final Solution: $f'(x) = (2x - 7)(3x + 5)^2 (30x - 43)$


More Questions from Class 11 Mathematics Limits and Derivatives EXERCISE 13.3


CBSE Solutions for Class 11 Mathematics Limits and Derivatives


Chapters in CBSE - Class 11 Mathematics


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