Differentiation (Part-38) | S N De

Differentiation (Part-33) | S N Dey Class 12 Mathematics Solutions

Key Facts About Differentiation

• Definition: Differentiation measures the instantaneous rate at which a function changes with respect to its input variable, representing the slope of the function curve at any point.
• Power Rule: Used to differentiate polynomial functions: \(\frac{d}{dx}(x^n) = n x^{n-1}\).
• Chain Rule: Used for composite functions: \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\).
• Second Derivative: Represents the rate of change of the first derivative (concavity/slope modification).
• Calculus Relationship: Differentiation is the inverse operation of integration; together they form the core foundation of calculus.

Below are step-by-step solutions for exercises from S N Dey Mathematics (Class 12).

12. If \( f(x) = \frac{1}{x} \), find the value of \( \theta \) in the mean value theorem \( f(x+h) = f(x) + h f'(x+\theta h) \), where \( 0 < \theta < 1 \).

Solution:

We are given:

$$ f(x) = \frac{1}{x} = x^{-1} \quad \rightarrow (1) $$

$$ \implies f'(x) = -x^{-2} = -\frac{1}{x^2} \quad \rightarrow (2) $$

According to the Mean Value Theorem statement:

$$ f(x+h) = f(x) + h f'(x+\theta h) $$

Substituting (1) and (2) into the equation:

$$ \frac{1}{x+h} = \frac{1}{x} + h \left[ -\frac{1}{(x+\theta h)^2} \right] $$

$$ \implies \frac{1}{x+h} - \frac{1}{x} = -\frac{h}{(x+\theta h)^2} $$

$$ \implies \frac{x - (x+h)}{x(x+h)} = -\frac{h}{(x+\theta h)^2} $$

$$ \implies \frac{-h}{x(x+h)} = \frac{-h}{(x+\theta h)^2} \quad (\text{since } h \neq 0) $$

Equating denominators:

$$ x(x+h) = (x+\theta h)^2 $$

$$ \implies x^2 + xh = x^2 + 2x h \theta + \theta^2 h^2 $$

$$ \implies \theta^2 h^2 + 2x h \theta - xh = 0 $$

Applying the quadratic formula to solve for \( \theta \):

$$ \theta = \frac{-2xh \pm \sqrt{(2xh)^2 - 4(h^2)(-xh)}}{2h^2} $$

$$ \implies \theta = \frac{-2xh \pm \sqrt{4x^2h^2 + 4xh^3}}{2h^2} = \frac{-2xh \pm 2h\sqrt{x^2+xh}}{2h^2} $$

$$ \implies \theta = \frac{-x \pm \sqrt{x^2+xh}}{h} $$

Since \( 0 < \theta < 1 \), taking the positive root:

\( \therefore \theta = \frac{\sqrt{x^2+xh} - x}{h} \)

13. Using Lagrange's mean value theorem, find a point on the curve \( y = \sqrt{x-2} \) defined in the interval \( [2, 3] \) where the tangent to the curve is parallel to the chord joining the endpoints of the curve.

Solution:

Let \( y = f(x) = \sqrt{x-2} \quad \rightarrow (1) \)

$$ \implies f'(x) = \frac{d}{dx}(x-2)^{1/2} = \frac{1}{2\sqrt{x-2}} \quad \rightarrow (2) $$

For \( f(x) \) to be defined in \( [2, 3] \), we require \( x - 2 \ge 0 \implies x \ge 2 \).

The function \( f(x) \) is continuous in \( [2, 3] \) and differentiable in \( (2, 3) \). Therefore, by Lagrange's Mean Value Theorem, there exists at least one value \( c \in (2, 3) \) such that:

$$ f'(c) = \frac{f(3) - f(2)}{3 - 2} $$

$$ \implies \frac{1}{2\sqrt{c-2}} = \frac{\sqrt{3-2} - \sqrt{2-2}}{1} $$

$$ \implies \frac{1}{2\sqrt{c-2}} = 1 $$

Squaring both sides:

$$ \frac{1}{4(c-2)} = 1 \implies 4c - 8 = 1 \implies 4c = 9 \implies c = \frac{9}{4} $$

Now, find the corresponding \( y \)-coordinate at \( c = \frac{9}{4} \):

$$ f\left(\frac{9}{4}\right) = \sqrt{\frac{9}{4} - 2} = \sqrt{\frac{9-8}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} $$

\( \therefore \) The required point on the curve is \( \left(\frac{9}{4}, \frac{1}{2}\right) \).

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14. Using Lagrange's mean value theorem, find a point on the curve \( y = x^2 \) where the tangent to the curve is parallel to the line joining the points \( (1, 1) \) and \( (2, 4) \).

Solution:

Let the given curve be \( y = f(x) = x^2 \quad \rightarrow (1) \)

$$ \implies f'(x) = 2x \quad \rightarrow (2) $$

Since \( f(x) \) is a polynomial function, it is continuous and differentiable everywhere in any finite interval. Consider the closed interval \( [1, 2] \).

By Lagrange's Mean Value Theorem, there exists a point \( c \in (1, 2) \) such that:

$$ f'(c) = \frac{f(2) - f(1)}{2 - 1} $$

$$ \implies 2c = \frac{2^2 - 1^2}{1} = \frac{4 - 1}{1} = 3 $$

$$ \implies c = \frac{3}{2} $$

Calculating the corresponding \( y \)-value \( f(c) \):

$$ f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 = \frac{9}{4} $$

\( \therefore \) The required point on the curve is \( \left(\frac{3}{2}, \frac{9}{4}\right) \).