Math Notes For Class 12 Chapter 2 Ex 2.4 Solved

What are the Core Formulas and Definitions for this Exercise?

In Class 12 Differentiation (Chapter 2, Exercise 2.4), we transition from basic derivative rules to composite, implicit, and parametric systems. Mastery of these rules is critical for solving higher-order differential equations and physics applications.

1. The Chain Rule for Composite Functions

If $y$ is a differentiable function of $u$, and $u$ is a differentiable function of $x$, then $y$ is a differentiable function of $x$, defined as:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

2. Implicit Differentiation Technique

When an equation expresses $x$ and $y$ implicitly in the form $f(x, y) = 0$, $y$ cannot be easily isolated. We differentiate both sides of the equation with respect to $x$, applying the product, quotient, or power rules, and treating $y$ as a function of $x$ (so $\frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx}$). Finally, we collect all terms containing $\frac{dy}{dx}$ on one side to solve for it.

3. Derivatives of Parametric Equations

When variables $x$ and $y$ are both given in terms of a third independent variable (parameter) $t$, such that $x = f(t)$ and $y = g(t)$, the derivative $\frac{dy}{dx}$ is evaluated using the parameter $t$ via the Chain Rule:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \quad \text{where } \frac{dx}{dt} \neq 0$$

4. Differentiation of One Function with Respect to Another Function

To differentiate $u = f(x)$ with respect to $v = g(x)$, we calculate:

$$\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}}$$

Rule / Method	Mathematical Formula	Key Application
Chain Rule	$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$	Composite algebraic or radical expressions
Implicit Function Rule	$\frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx}$	Equations where $x$ and $y$ cannot be separated
Parametric Derivative	$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$	Curves defined by parametric equations $x(t), y(t)$
Function w.r.t Function	$\frac{du}{dv} = \frac{du/dx}{dv/dx}$	Rates of change between two mathematical models

How to Solve All Exercise Questions Step-by-Step?

Question 1: Find $\frac{dy}{dx}$ by making suitable substitutions for $y = \sqrt{\frac{1-x}{1+x}}$

Solution:

Let $u = \frac{1-x}{1+x}$, then $y = \sqrt{u} = u^{1/2}$.

First, differentiate $y$ with respect to $u$:

$$\frac{dy}{du} = \frac{d}{du}(u^{1/2}) = \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}}$$

Next, differentiate $u$ with respect to $x$ using the Quotient Rule:

$$\frac{du}{dx} = \frac{d}{dx}\left(\frac{1-x}{1+x}\right) = \frac{(1+x)\frac{d}{dx}(1-x) - (1-x)\frac{d}{dx}(1+x)}{(1+x)^2}$$ $$\frac{du}{dx} = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1 - x - 1 + x}{(1+x)^2} = \frac{-2}{(1+x)^2}$$

Now, apply the Chain Rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

$$\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \frac{-2}{(1+x)^2}$$ $$\frac{dy}{dx} = \frac{\sqrt{1+x}}{\sqrt{1-x}} \cdot \frac{-1}{(1+x)^2} = \frac{-1}{\sqrt{1-x} (1+x)^{3/2}}$$ ---

Question 2: Find $\frac{dy}{dx}$ for the implicit relation $3x^2 + 4xy - 5y^2 + 2x - 3y + 7 = 0$

Solution:

Differentiate both sides of the equation with respect to $x$:

$$\frac{d}{dx}(3x^2) + \frac{d}{dx}(4xy) - \frac{d}{dx}(5y^2) + \frac{d}{dx}(2x) - \frac{d}{dx}(3y) + \frac{d}{dx}(7) = 0$$

Apply the Product Rule to $4xy$ and Power/Chain Rules to $y$ terms:

$$6x + 4\left(y + x\frac{dy}{dx}\right) - 10y\frac{dy}{dx} + 2 - 3\frac{dy}{dx} + 0 = 0$$ $$6x + 4y + 4x\frac{dy}{dx} - 10y\frac{dy}{dx} + 2 - 3\frac{dy}{dx} = 0$$

Group all terms containing $\frac{dy}{dx}$ on the left side:

$$\frac{dy}{dx}(4x - 10y - 3) = -6x - 4y - 2$$ $$\frac{dy}{dx} = \frac{-(6x + 4y + 2)}{4x - 10y - 3} = \frac{6x + 4y + 2}{10y - 4x + 3}$$ ---

Question 3: If $x = a\left(\frac{1-t^2}{1+t^2}\right)$ and $y = \frac{2bt}{1+t^2}$, find $\frac{dy}{dx}$

Solution:

Differentiate $x$ with respect to $t$ using Quotient Rule:

$$\frac{dx}{dt} = a \cdot \frac{(1+t^2)\frac{d}{dt}(1-t^2) - (1-t^2)\frac{d}{dt}(1+t^2)}{(1+t^2)^2}$$ $$\frac{dx}{dt} = a \cdot \frac{(1+t^2)(-2t) - (1-t^2)(2t)}{(1+t^2)^2} = a \cdot \frac{-2t - 2t^3 - 2t + 2t^3}{(1+t^2)^2} = \frac{-4at}{(1+t^2)^2}$$

Differentiate $y$ with respect to $t$ using Quotient Rule:

$$\frac{dy}{dt} = 2b \cdot \frac{(1+t^2)\frac{d}{dt}(t) - t\frac{d}{dt}(1+t^2)}{(1+t^2)^2}$$ $$\frac{dy}{dt} = 2b \cdot \frac{(1+t^2)(1) - t(2t)}{(1+t^2)^2} = 2b \cdot \frac{1+t^2-2t^2}{(1+t^2)^2} = \frac{2b(1-t^2)}{(1+t^2)^2}$$

Using the Parametric Chain Rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

$$\frac{dy}{dx} = \frac{\frac{2b(1-t^2)}{(1+t^2)^2}}{\frac{-4at}{(1+t^2)^2}} = \frac{2b(1-t^2)}{-4at} = \frac{-b(1-t^2)}{2at}$$ ---

Question 4: Differentiate $\left(x^2 + \frac{1}{x^2}\right)$ with respect to $\left(x - \frac{1}{x}\right)$

Solution:

Let $u = x^2 + x^{-2}$ and $v = x - x^{-1}$. We need to find $\frac{du}{dv}$.

First, differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = 2x - 2x^{-3} = 2x - \frac{2}{x^3} = \frac{2(x^4 - 1)}{x^3}$$

Next, differentiate $v$ with respect to $x$:

$$\frac{dv}{dx} = 1 - (-1)x^{-2} = 1 + \frac{1}{x^2} = \frac{x^2 + 1}{x^2}$$

Now, calculate $\frac{du}{dv} = \frac{du/dx}{dv/dx}$:

$$\frac{du}{dv} = \frac{\frac{2(x^4 - 1)}{x^3}}{\frac{x^2 + 1}{x^2}} = \frac{2(x^2 - 1)(x^2 + 1)}{x^3} \cdot \frac{x^2}{x^2 + 1}$$ $$\frac{du}{dv} = \frac{2(x^2 - 1)}{x} = 2\left(x - \frac{1}{x}\right)$$

Interactive Practice Quiz: Test Your Understanding (Clickable MCQs)

Frequently Asked Questions: What are Common Student Errors in this Exercise?

Q1: Why do students lose marks when differentiating $y^2$ or $y^3$ implicitly?
Ans: The single most common mistake in Pakistani board exams (FBISE/BISE) is writing $\frac{d}{dx}[y^2] = 2y$ instead of $2y \frac{dy}{dx}$. Students forget that $y$ is an implicit function of $x$. Whenever you differentiate a term containing $y$ with respect to $x$, you must attach the inner derivative multiplier $\frac{dy}{dx}$ via the Chain Rule.

Q2: How do students make mistakes in parametric equation derivatives?
Ans: Students frequently write $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt}$ instead of using division $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. Always remember that by the chain rule, $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy/dt}{dx/dt}$. Double-check that you invert $\frac{dx}{dt}$ when multiplying.

Q3: What is the correct approach when differentiating one function $u(x)$ w.r.t another function $v(x)$?
Ans: Do not attempt to isolate $x$ algebraically to substitute $v$ directly into $u$, as this often leads to complicated expressions. Instead, explicitly declare $u = f(x)$ and $v = g(x)$, calculate their independent derivatives $\frac{du}{dx}$ and $\frac{dv}{dx}$ separately, and then divide $\frac{du}{dv} = \frac{du/dx}{dv/dx}$. This structured method avoids heavy algebraic errors during board exams.