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Q17(i):
Differentiate $(x^2 - 5x + 8)(x^3 + 7x + 9)$ in three ways mentioned below: (i) by using product rule
Solution :
Step 1: Establishing the Product Rule Framework
To differentiate the product of two functions $u(x)$ and $v(x)$, we employ the Product Rule of differentiation, defined as follows [Per Leibniz's Rule for Differentiation]:
$\frac{d}{dx}[u(x) \cdot v(x)] = u(x) \cdot \frac{d}{dx}[v(x)] + v(x) \cdot \frac{d}{dx}[u(x)]$
Let the given function $y$ be defined as:
$y = (x^2 - 5x + 8)(x^3 + 7x + 9)$
Where:
- $u(x) = x^2 - 5x + 8$
- $v(x) = x^3 + 7x + 9$
Step 2: Calculating Individual Derivatives
Applying the Power Rule $\frac{d}{dx}(x^n) = nx^{n-1}$ [By the Power Rule of Calculus]:
$u'(x) = \frac{d}{dx}(x^2 - 5x + 8) = 2x - 5$
$v'(x) = \frac{d}{dx}(x^3 + 7x + 9) = 3x^2 + 7$
Step 3: Substitution into the Product Rule Formula
Substituting $u, v, u',$ and $v'$ into the product rule expression:
$\frac{dy}{dx} = (x^2 - 5x + 8)(3x^2 + 7) + (x^3 + 7x + 9)(2x - 5)$
Step 4: Algebraic Expansion
Expand the terms individually to simplify the expression [By the Distributive Property of Multiplication over Addition]:
Term 1 Expansion: $(x^2 - 5x + 8)(3x^2 + 7) = 3x^4 + 7x^2 - 15x^3 - 35x + 24x^2 + 56 = 3x^4 - 15x^3 + 31x^2 - 35x + 56$
Term 2 Expansion: $(x^3 + 7x + 9)(2x - 5) = 2x^4 - 5x^3 + 14x^2 - 35x + 18x - 45 = 2x^4 - 5x^3 + 14x^2 - 17x - 45$
Step 5: Summation of Terms
Group and combine like terms:
$\frac{dy}{dx} = (3x^4 + 2x^4) + (-15x^3 - 5x^3) + (31x^2 + 14x^2) + (-35x - 17x) + (56 - 45)$
$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$
Final Solution: \frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11
More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.5
- Q1: Differentiate the functions w.r.t. $x$. $\cos x \cdot \cos 2x \cdot \cos 3x$
- Q2: Differentiate the functions w.r.t. $x$. $\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$
- Q3: Differentiate the functions w.r.t. $x$. $(\log x)^{\cos x}$
- Q4: Differentiate the functions w.r.t. $x$. $x^x - 2^{\sin x}$
- Q5: Differentiate the functions w.r.t. $x$. $(x + 3)^2 \cdot (x + 4)^3 \cdot (x + 5)^4$
- Q6: Differentiate the functions w.r.t. $x$. $\left(x + \frac{1}{x}\right)^x + x^{\left(1 + \frac{1}{x}\right)}$
- Q7: Differentiate the functions w.r.t. $x$. $(\log x)^x + x^{\log x}$
- Q8: Differentiate the functions w.r.t. $x$. $(\sin x)^x + \sin^{-1} \sqrt{x}$
- Q9: Differentiate the functions w.r.t. $x$. $x^{\sin x} + (\sin x)^{\cos x}$
- Q10: Differentiate the functions w.r.t. $x$. $x^{x \cos x} + \frac{x^2+1}{x^2-1}$
- Q11: Differentiate the functions w.r.t. $x$. $(x \cos x)^x + (x \sin x)^{\frac{1}{x}}$
- Q12: Find $\frac{dy}{dx}$ of the functions $x^y + y^x = 1$
- Q13: Find $\frac{dy}{dx}$ of the functions $y^x = x^y$
- Q14: Find $\frac{dy}{dx}$ of the functions $(\cos x)^y = (\cos y)^x$
- Q15: Find $\frac{dy}{dx}$ of the functions $xy = e^{(x - y)}$
- Q16: Find the derivative of the function given by $f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)$ and hence find $f'(1)$.
- Q17(ii): Differentiate $(x^2 - 5x + 8)(x^3 + 7x + 9)$ in three ways mentioned below: (ii) by expanding the product to obtain a single polynomial.
- Q17(iii): Differentiate $(x^2 - 5x + 8)(x^3 + 7x + 9)$ in three ways mentioned below: (iii) by logarithmic differentiation. Do they all give the same answer?
- Q18: If $u, v$ and $w$ are functions of $x$, then show that $\frac{d}{dx} (u \cdot v \cdot w) = \frac{du}{dx} v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \frac{dw}{dx}$ in two ways - first by repeated application of product rule, second by logarithmic differentiation.
CBSE Solutions for Class 12 Mathematics Continuity and Differentiability
- Continuity and Differentiability Class 12 EXERCISE 5.1
- Continuity and Differentiability Class 12 EXERCISE 5.2
- Continuity and Differentiability Class 12 EXERCISE 5.3
- Continuity and Differentiability Class 12 EXERCISE 5.4
- Continuity and Differentiability Class 12 EXERCISE 5.6
- Continuity and Differentiability Class 12 EXERCISE 5.7
Chapters in CBSE - Class 12 Mathematics
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