Math Notes For Class 12 Chapter 2 Ex 2.3 Solved & Quiz

Question 1: Differentiate with respect to $x$: $$y = x^4 + 2x^3 + x^2$$

Solution:
Let $y = x^4 + 2x^3 + x^2$.
Differentiating both sides with respect to $x$: $$\frac{dy}{dx} = \frac{d}{dx}(x^4 + 2x^3 + x^2)$$ Applying the Sum Rule and Constant Multiple Rule: $$\frac{dy}{dx} = \frac{d}{dx}(x^4) + 2 \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2)$$ Applying the Power Rule $\frac{d}{dx}(x^n) = n x^{n-1}$: $$\frac{dy}{dx} = 4x^{4-1} + 2(3x^{3-1}) + 2x^{2-1}$$ $$\frac{dy}{dx} = 4x^3 + 6x^2 + 2x$$ Factoring out $2x$: $$\frac{dy}{dx} = 2x(2x^2 + 3x + 1)$$
Answer: $$\frac{dy}{dx} = 2x(2x+1)(x+1)$$

Question 2: Differentiate with respect to $x$: $$y = x^{-3} + 2x^{-2} + 3$$

Solution:
Let $y = x^{-3} + 2x^{-2} + 3$.
Differentiating both sides with respect to $x$: $$\frac{dy}{dx} = \frac{d}{dx}(x^{-3}) + 2\frac{d}{dx}(x^{-2}) + \frac{d}{dx}(3)$$ Using the Power Rule and Constant Rule: $$\frac{dy}{dx} = (-3)x^{-3-1} + 2(-2)x^{-2-1} + 0$$ $$\frac{dy}{dx} = -3x^{-4} - 4x^{-3}$$ Converting negative exponents into positive denominators: $$\frac{dy}{dx} = -\frac{3}{x^4} - \frac{4}{x^3}$$ Taking $x^4$ as the common denominator: $$\frac{dy}{dx} = \frac{-3 - 4x}{x^4} = -\frac{3+4x}{x^4}$$
Answer: $$\frac{dy}{dx} = -\frac{4x+3}{x^4}$$

Question 3: Differentiate with respect to $x$: $$y = \frac{a+x}{a-x}$$

Solution:
Let $y = \frac{a+x}{a-x}$, where $a$ is a constant.
Applying the Quotient Rule $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v u' - u v'}{v^2}$ with $u = a+x$ and $v = a-x$: $$\frac{dy}{dx} = \frac{(a-x) \frac{d}{dx}(a+x) - (a+x) \frac{d}{dx}(a-x)}{(a-x)^2}$$ Differentiating the numerators: $$\frac{d}{dx}(a+x) = 0 + 1 = 1$$ $$\frac{d}{dx}(a-x) = 0 - 1 = -1$$ Substitute these derivatives back into the expression: $$\frac{dy}{dx} = \frac{(a-x)(1) - (a+x)(-1)}{(a-x)^2}$$ Simplify the numerator: $$\frac{dy}{dx} = \frac{a - x + a + x}{(a-x)^2} = \frac{2a}{(a-x)^2}$$
Answer: $$\frac{dy}{dx} = \frac{2a}{(a-x)^2}$$

Question 4: Differentiate with respect to $x$: $$y = \frac{2x-3}{2x+1}$$

Solution:
Let $y = \frac{2x-3}{2x+1}$.
Using the Quotient Rule with $u = 2x-3$ and $v = 2x+1$: $$\frac{dy}{dx} = \frac{(2x+1) \frac{d}{dx}(2x-3) - (2x-3) \frac{d}{dx}(2x+1)}{(2x+1)^2}$$ Calculate individual derivatives: $$\frac{d}{dx}(2x-3) = 2(1) - 0 = 2$$ $$\frac{d}{dx}(2x+1) = 2(1) + 0 = 2$$ Substituting values into the quotient formula: $$\frac{dy}{dx} = \frac{(2x+1)(2) - (2x-3)(2)}{(2x+1)^2}$$ Expanding the terms in the numerator: $$\frac{dy}{dx} = \frac{(4x + 2) - (4x - 6)}{(2x+1)^2}$$ $$\frac{dy}{dx} = \frac{4x + 2 - 4x + 6}{(2x+1)^2} = \frac{8}{(2x+1)^2}$$
Answer: $$\frac{dy}{dx} = \frac{8}{(2x+1)^2}$$

Question 5: Differentiate with respect to $x$: $$y = (x-5)(1-x)$$

Solution:
Method 1 (Algebraic Expansion):
Expand the expression prior to differentiation: $$y = x(1) - x(x) - 5(1) - 5(-x)$$ $$y = x - x^2 - 5 + 5x = -x^2 + 6x - 5$$ Differentiating term-by-term with respect to $x$: $$\frac{dy}{dx} = \frac{d}{dx}(-x^2) + \frac{d}{dx}(6x) - \frac{d}{dx}(5)$$ $$\frac{dy}{dx} = -2x + 6 - 0 = 6 - 2x = 2(3-x)$$
Method 2 (Product Rule):
Let $u = x-5$ and $v = 1-x$. $$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = (x-5)\frac{d}{dx}(1-x) + (1-x)\frac{d}{dx}(x-5)$$ $$\frac{dy}{dx} = (x-5)(-1) + (1-x)(1) = -x + 5 + 1 - x = 6 - 2x$$
Answer: $$\frac{dy}{dx} = 2(3-x)$$

Question 6: Differentiate with respect to $x$: $$y = \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2$$

Solution:
First, expand the algebraic square $(a-b)^2 = a^2 - 2ab + b^2$: $$y = (\sqrt{x})^2 - 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) + \left(\frac{1}{\sqrt{x}}\right)^2$$ $$y = x - 2 + \frac{1}{x} = x - 2 + x^{-1}$$ Now, differentiate each term with respect to $x$: $$\frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(2) + \frac{d}{dx}(x^{-1})$$ Applying power rule: $$\frac{d}{dx}(x) = 1$$ $$\frac{d}{dx}(2) = 0$$ $$\frac{d}{dx}(x^{-1}) = (-1)x^{-1-1} = -x^{-2} = -\frac{1}{x^2}$$ Combining the result: $$\frac{dy}{dx} = 1 - 0 - \frac{1}{x^2} = 1 - \frac{1}{x^2}$$ Combining into a single fraction: $$\frac{dy}{dx} = \frac{x^2 - 1}{x^2}$$
Answer: $$\frac{dy}{dx} = \frac{x^2 - 1}{x^2}$$