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Key Points About Differentiation in Mathematics
- Rate of Change: Differentiation allows us to calculate the instantaneous rate at which a function changes with respect to its independent variable at any given point.
- Tangent Lines: The derivative gives the slope of the tangent line to the curve at a specific point, representing the best linear approximation near that point.
- Higher Order Derivatives: Applied repeatedly to analyze curvature and concavity (the second derivative represents the rate of change of slope).
- Optimization: Critical points (where \(f'(x) = 0\)) allow determination of local maxima and minima for real-world optimization problems.
- Differentiation Rules: Systematic rules (power, product, quotient, chain rule) enable efficient differentiation of complex functions.
Solutions: S N Dey Class 12 Exercises
Solution:
The trigonometric function \( f(x) = \cos^2 x \) is continuous in \( [-\pi/4, \pi/4] \) and differentiable in \( (-\pi/4, \pi/4) \).
Checking boundary conditions:
$$ f(-\pi/4) = \cos^2(-\pi/4) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} $$
$$ f(\pi/4) = \cos^2(\pi/4) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} $$
$$ \implies f(-\pi/4) = f(\pi/4) $$
Since all conditions of Rolle's theorem are satisfied, there exists at least one value \( c \in (-\pi/4, \pi/4) \) such that \( f'(c) = 0 \).
Differentiating \( f(x) \):
$$ f'(x) = \frac{d}{dx}(\cos^2 x) = 2\cos x \cdot (-\sin x) = -\sin 2x $$
Setting \( f'(c) = 0 \):
$$ -\sin 2c = 0 \implies \sin 2c = 0 = \sin(n\pi) \quad [n \in \mathbb{Z}] $$
$$ 2c = n\pi \implies c = \frac{n\pi}{2} $$
For \( n = 0 \), \( c = 0 \in (-\pi/4, \pi/4) \).
Solution:
Being a polynomial function, \( f(x) \) is continuous in \( [-3, 4] \) and differentiable in \( (-3, 4) \).
Checking boundary values:
$$ f(-3) = (-3)^2 - (-3) - 12 = 9 + 3 - 12 = 0 $$
$$ f(4) = (4)^2 - 4 - 12 = 16 - 4 - 12 = 0 $$
$$ \implies f(-3) = f(4) = 0 $$
By Rolle's theorem, there exists \( c \in (-3, 4) \) such that \( f'(c) = 0 \).
Differentiating \( f(x) \):
$$ f'(x) = \frac{d}{dx}(x^2 - x - 12) = 2x - 1 $$
Setting \( f'(c) = 0 \):
$$ 2c - 1 = 0 \implies c = \frac{1}{2} $$
Solution:
As a polynomial function, \( f(x) \) is continuous in \( [2, 3] \) and differentiable in \( (2, 3) \).
Checking boundary values:
$$ f(2) = 2^2 - 5(2) + 6 = 4 - 10 + 6 = 0 $$
$$ f(3) = 3^2 - 5(3) + 6 = 9 - 15 + 6 = 0 $$
$$ \implies f(2) = f(3) = 0 $$
By Rolle's theorem, there exists \( c \in (2, 3) \) such that \( f'(c) = 0 \).
Differentiating \( f(x) \):
$$ f'(x) = 2x - 5 $$
Setting \( f'(c) = 0 \):
$$ 2c - 5 = 0 \implies c = \frac{5}{2} $$
Solution:
The trigonometric function \( f(x) = \cos x \) is continuous in \( [-\pi/2, \pi/2] \) and differentiable in \( (-\pi/2, \pi/2) \).
Checking boundary values:
$$ f(-\pi/2) = \cos(-\pi/2) = 0 $$
$$ f(\pi/2) = \cos(\pi/2) = 0 $$
$$ \implies f(-\pi/2) = f(\pi/2) = 0 $$
By Rolle's theorem, there exists \( c \in (-\pi/2, \pi/2) \) such that \( f'(c) = 0 \).
Differentiating \( f(x) \):
$$ f'(x) = -\sin x $$
Setting \( f'(c) = 0 \):
$$ -\sin c = 0 \implies \sin c = 0 = \sin(n\pi) \quad [n \in \mathbb{Z}] $$
For \( n = 0 \), \( c = 0 \in (-\pi/2, \pi/2) \).
Solution:
The trigonometric function \( f(x) = \sin 3x \) is continuous in \( [0, \pi] \) and differentiable in \( (0, \pi) \).
Checking boundary values:
$$ f(0) = \sin 0 = 0 $$
$$ f(\pi) = \sin(3\pi) = 0 $$
$$ \implies f(0) = f(\pi) = 0 $$
By Rolle's theorem, there exists \( c \in (0, \pi) \) such that \( f'(c) = 0 \).
Differentiating \( f(x) \):
$$ f'(x) = 3\cos 3x $$
Setting \( f'(c) = 0 \):
$$ 3\cos 3c = 0 \implies \cos 3c = 0 \implies 3c = (2n + 1)\frac{\pi}{2} \quad [n \in \mathbb{Z}] $$
For \( n = 0 \):
$$ 3c = \frac{\pi}{2} \implies c = \frac{\pi}{6} $$
Solution:
Since \( e^{-x} \) and \( \sin x \) are continuous on \( \mathbb{R} \), their product \( f(x) = e^{-x}\sin x \) is continuous in \( [0, \pi] \).
Differentiating using product rule:
$$ f'(x) = \frac{d}{dx}(e^{-x})\cdot \sin x + e^{-x}\cdot \frac{d}{dx}(\sin x) $$
$$ f'(x) = -e^{-x}\sin x + e^{-x}\cos x = e^{-x}(\cos x - \sin x) $$
Clearly, \( f'(x) \) exists for all \( x \in (0, \pi) \).
Checking boundary values:
$$ f(0) = e^0 \sin 0 = 0 \quad \text{and} \quad f(\pi) = e^{-\pi}\sin \pi = 0 $$
$$ \implies f(0) = f(\pi) = 0 $$
Setting \( f'(c) = 0 \):
$$ e^{-c}(\cos c - \sin c) = 0 $$
Since \( e^{-c} \neq 0 \):
$$ \cos c - \sin c = 0 \implies \tan c = 1 \implies c = \frac{\pi}{4} $$
Solution:
Being a polynomial function, \( f(x) \) is continuous in \( [1, 3] \) and differentiable in \( (1, 3) \).
Checking boundary values:
$$ f(1) = 1^2 - 4(1) + 3 = 1 - 4 + 3 = 0 $$
$$ f(3) = 3^2 - 4(3) + 3 = 9 - 12 + 3 = 0 $$
$$ \implies f(1) = f(3) = 0 $$
By Rolle's theorem, there exists \( c \in (1, 3) \) such that \( f'(c) = 0 \).
Differentiating \( f(x) \):
$$ f'(x) = 2x - 4 $$
Setting \( f'(c) = 0 \):
$$ 2c - 4 = 0 \implies c = 2 $$
Solution:
As the difference of two continuous trigonometric functions, \( f(x) \) is continuous in \( [0, 2\pi] \).
Differentiating \( f(x) \):
$$ f'(x) = \cos x - 2\cos 2x $$
Clearly, \( f'(x) \) exists for all \( x \in (0, 2\pi) \).
Checking boundary values:
$$ f(0) = \sin 0 - \sin 0 = 0 \quad \text{and} \quad f(2\pi) = \sin(2\pi) - \sin(4\pi) = 0 $$
$$ \implies f(0) = f(2\pi) = 0 $$
Setting \( f'(c) = 0 \):
$$ \cos c - 2\cos 2c = 0 \implies \cos c - 2(2\cos^2 c - 1) = 0 $$
$$ 4\cos^2 c - \cos c - 2 = 0 $$
Applying quadratic formula for \( \cos c \):
$$ \cos c = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-2)}}{2(4)} = \frac{1 \pm \sqrt{33}}{8} $$
Since \( \frac{1 + \sqrt{33}}{8} \approx 0.843 \in [-1, 1] \) and \( \frac{1 - \sqrt{33}}{8} \approx -0.593 \in [-1, 1] \), both values are valid within the domain.
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