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Math Notes For Class 12 Chapter 2 Ex 2.4 Solved
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Math Notes For Class 12 Chapter 2 Ex 2.4 Solved
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.
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:
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$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.
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:
$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \quad \text{where } \frac{dx}{dt} \neq 0$$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 |
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 $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:
$$\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \frac{-2}{(1+x)^2}$$ $$\frac{dy}{dx} = \frac{\sqrt{1+x}}{\sqrt{1-x}} \cdot \frac{-1}{(1+x)^2} = \frac{-1}{\sqrt{1-x} (1+x)^{3/2}}$$ ---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:
$$\frac{dy}{dx}(4x - 10y - 3) = -6x - 4y - 2$$ $$\frac{dy}{dx} = \frac{-(6x + 4y + 2)}{4x - 10y - 3} = \frac{6x + 4y + 2}{10y - 4x + 3}$$ ---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 $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$$\frac{dy}{dx} = \frac{\frac{2b(1-t^2)}{(1+t^2)^2}}{\frac{-4at}{(1+t^2)^2}} = \frac{2b(1-t^2)}{-4at} = \frac{-b(1-t^2)}{2at}$$ ---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 $\frac{du}{dv} = \frac{du/dx}{dv/dx}$:
$$\frac{du}{dv} = \frac{\frac{2(x^4 - 1)}{x^3}}{\frac{x^2 + 1}{x^2}} = \frac{2(x^2 - 1)(x^2 + 1)}{x^3} \cdot \frac{x^2}{x^2 + 1}$$ $$\frac{du}{dv} = \frac{2(x^2 - 1)}{x} = 2\left(x - \frac{1}{x}\right)$$Q1: If $y = f(u)$ and $u = g(x)$, what is the correct formula for $\frac{dy}{dx}$?
Q2: What is the derivative of $y^3$ with respect to $x$?
Q3: If $x = a t^2$ and $y = 2at$, what is $\frac{dy}{dx}$?
Q4: For the circle equation $x^2 + y^2 = r^2$, what is $\frac{dy}{dx}$?
Q5: What is the derivative of $x^3$ with respect to $x^2$?
Q6: If $x = \cos(\theta)$ and $y = \sin(\theta)$, what is $\frac{dy}{dx}$?
Q7: When applying implicit differentiation to the term $xy$, which derivative rule is used?
Q8: What is $\frac{dy}{dx}$ if $xy = 1$?
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.