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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?
Solution :
Step 1: Establishing the Logarithmic Framework
Given the function $y = (x^2 - 5x + 8)(x^3 + 7x + 9)$, we apply the natural logarithm to both sides to simplify the product into a sum of logarithms.
$\ln(y) = \ln((x^2 - 5x + 8)(x^3 + 7x + 9))$
[By the property of logarithms: $\ln(ab) = \ln(a) + \ln(b)$]
$\ln(y) = \ln(x^2 - 5x + 8) + \ln(x^3 + 7x + 9)$
Step 2: Differentiating with respect to $x$
Differentiate both sides with respect to $x$ using the chain rule for the logarithmic terms, $\frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx}$.
$\frac{d}{dx}(\ln(y)) = \frac{d}{dx}(\ln(x^2 - 5x + 8)) + \frac{d}{dx}(\ln(x^3 + 7x + 9))$
$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x^2 - 5x + 8} \cdot \frac{d}{dx}(x^2 - 5x + 8) + \frac{1}{x^3 + 7x + 9} \cdot \frac{d}{dx}(x^3 + 7x + 9)$
Step 3: Evaluating Derivatives and Isolating $\frac{dy}{dx}$
Calculate the inner derivatives:
$\frac{d}{dx}(x^2 - 5x + 8) = 2x - 5$
$\frac{d}{dx}(x^3 + 7x + 9) = 3x^2 + 7$
Substitute these into the equation:
$\frac{1}{y} \frac{dy}{dx} = \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9}$
Isolate $\frac{dy}{dx}$ by multiplying by $y$:
$\frac{dy}{dx} = (x^2 - 5x + 8)(x^3 + 7x + 9) \left[ \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right]$
Step 4: Final Algebraic Expansion
Multiply $y$ back into the brackets to reach the standard polynomial form:
$\frac{dy}{dx} = (2x - 5)(x^3 + 7x + 9) + (3x^2 + 7)(x^2 - 5x + 8)$
Expanding the terms:
$(2x^4 + 14x^2 + 18x - 5x^3 - 35x - 45) + (3x^4 - 15x^3 + 24x^2 + 7x^2 - 35x + 56)$
Grouping like terms:
$\frac{dy}{dx} = (2x^4 + 3x^4) + (-5x^3 - 15x^3) + (14x^2 + 24x^2 + 7x^2) + (18x - 35x - 35x) + (-45 + 56)$
$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$
Step 5: Verification of Multi-Method Consistency
Whether using the Product Rule, direct polynomial expansion, or logarithmic differentiation, the fundamental laws of calculus dictate that the derivative of a differentiable function is unique [Uniqueness Theorem of Derivatives].
| Method | Consistency |
|---|---|
| Product Rule | Yields $5x^4 - 20x^3 + 45x^2 - 52x + 11$ |
| Expansion | Yields $5x^4 - 20x^3 + 45x^2 - 52x + 11$ |
| Logarithmic | Yields $5x^4 - 20x^3 + 45x^2 - 52x + 11$ |
Final Solution: Yes, all three methods yield the same result: $\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$
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(i): Differentiate $(x^2 - 5x + 8)(x^3 + 7x + 9)$ in three ways mentioned below: (i) by using product rule
- 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.
- 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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