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Math Notes For Class 12 Chapter 2 Ex 2.3 Solved & Quiz
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Math Notes For Class 12 Chapter 2 Ex 2.3 Solved & Quiz
Differentiation is the fundamental operation in calculus that measures the instantaneous rate of change of a function $y = f(x)$ with respect to its independent variable $x$. The derivative, denoted as $\frac{dy}{dx}$ or $f'(x)$, represents the gradient of the tangent line to the curve at any given point.
In Exercise 2.3 of Class 12 Mathematics (FSc Pre-Engineering), algebraic functions are differentiated using standard algebraic differentiation rules rather than the ab-initio (first principles) method. Mastering these rules ensures speed and numerical accuracy in subsequent higher-order calculus topics.
| Rule Name | Mathematical Notation / Formula | Description / Application |
|---|---|---|
| Constant Rule | $$\frac{d}{dx}(c) = 0$$ | The derivative of any constant value $c$ is zero. |
| Power Rule | $$\frac{d}{dx}(x^n) = n x^{n-1}$$ | Multiply by the power $n$ and reduce the exponent by $1$. |
| Constant Multiple Rule | $$\frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)]$$ | Constants factor out directly during differentiation. |
| Sum and Difference Rule | $$\frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)]$$ | Differentiate term-by-term across addition/subtraction. |
| Product Rule | $$\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$$ | First function times derivative of second plus second times derivative of first. |
| Quotient Rule | $$\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ | Denominator times derivative of numerator minus numerator times derivative of denominator, all over denominator squared. |
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: $$\frac{dy}{dx} = \frac{x^2 - 1}{x^2}$$
Q1: What is the derivative of $x^7$ with respect to $x$?
Q2: If $y = \frac{1}{x^3}$, what is $\frac{dy}{dx}$?
Q3: What is the derivative of a constant real number $k$?
Q4: According to the Quotient Rule, $\frac{d}{dx}\left[\frac{u}{v}\right]$ equals:
Q5: What is $\frac{d}{dx}(\sqrt{x})$?
Q6: Derivative of $y = (2x+3)^2$ with respect to $x$ is:
Q7: If $y = \frac{x}{x+1}$, what is $\frac{dy}{dx}$?