Continuity and Differentiability Ex 5.3 Class 12 NCERT

Understanding Implicit Differentiation & Inverse Trigonometric Derivatives

In implicit differentiation, we treat $y$ as an implicit function of $x$. Instead of solving the equation for $y$ first (which can be algebraically tedious or even impossible), we differentiate both sides of the equation term-by-term with respect to $x$. When doing this, any term containing $y$ requires the application of the Chain Rule. For example, the derivative of $y^2$ with respect to $x$ is not simply $2y$, but rather $2y \cdot \frac{dy}{dx}$.

The second half of Exercise 5.3 focuses on the derivatives of inverse trigonometric functions. While you can differentiate functions like $y = \sin^{-1}(x)$ directly using standard formulas (where $\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$), composite inverse trigonometric functions are much easier to solve using substitution. By substituting $x = \tan\theta$, $x = \sin\theta$, or other trigonometric variables, you can simplify complex algebraic expressions inside the inverse trigonometric functions using standard identities before finding the derivative.

Step-by-Step Method for Implicit Differentiation

1. Differentiate Both Sides — Apply the derivative operator $\frac{d}{dx}$ to every term on both the Left-Hand Side (LHS) and the Right-Hand Side (RHS) of the equation.
2. Apply the Chain and Product Rules — Use the product rule for terms involving both variables, such as $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$. Use the chain rule for terms involving only $y$, such as $\frac{d}{dx}(f(y)) = f'(y)\frac{dy}{dx}$.
3. Isolate the dy/dx Terms — Group all terms containing $\frac{dy}{dx}$ on one side of the equation (usually the LHS) and move all other terms to the opposite side.
4. Factor and Solve — Factor out $\frac{dy}{dx}$ from the grouped terms and divide to find the final algebraic expression for $\frac{dy}{dx}$.

Worked Solutions from Exercise 5.3 Concepts

• Example 1 (Implicit Differentiation): Find $\frac{dy}{dx}$ for the equation $x^2 + xy + y^2 = 100$. Step 1: Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(100)$ Step 2: Apply the product rule to $xy$ and chain rule to $y^2$: $2x + (x\frac{dy}{dx} + y\cdot 1) + 2y\frac{dy}{dx} = 0$ Step 3: Rearrange the terms to group $\frac{dy}{dx}$: $x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y$ Step 4: Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x + 2y) = -(2x + y)$ Step 5: Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$
• Example 2 (Inverse Trig Derivative): Find $\frac{dy}{dx}$ if $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$. Step 1: Use substitution to simplify. Let $x = \tan\theta$, which implies $\theta = \tan^{-1}x$. Step 2: Substitute into the equation: $y = \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)$ Step 3: Use the trigonometric identity $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$: $y = \sin^{-1}(\sin(2\theta)) = 2\theta$ Step 4: Replace $\theta$ back with $\tan^{-1}x$: $y = 2\tan^{-1}x$ Step 5: Differentiate with respect to $x$: $\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}x) = \frac{2}{1+x^2}$

CBSE Board Exam Tips & Common Mistakes

1. The Chain Rule Trap: The most common mistake is forgetting to write $\frac{dy}{dx}$ when differentiating a $y$-term. For instance, writing $\frac{d}{dx}(\cos y) = -\sin y$ instead of $-\sin y \cdot \frac{dy}{dx}$ is a classic error that will cost you full marks.
2. Product Rule Neglect: Terms like $xy$ or $x^2y$ must be expanded using the product rule. Always remember: $\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}$.
3. Domain Constraints: In inverse trigonometric questions, pay close attention to given domains (e.g., $0 < x < 1$). They dictate whether identities map directly within principal value branches. If the interval matches standard principal limits, you can directly simplify $\sin^{-1}(\sin\theta) = \theta$.

Practice Questions with Solutions

• Q: Find $\frac{dy}{dx}$ for $2x + 3y = \sin(y)$. A: Step 1: Differentiate both sides of the equation with respect to $x$: $\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$ Step 2: Apply the rules of differentiation: $2 + 3\frac{dy}{dx} = \cos(y) \cdot \frac{dy}{dx}$ Step 3: Collect terms containing $\frac{dy}{dx}$ on one side: $2 = \cos(y)\frac{dy}{dx} - 3\frac{dy}{dx}$ Step 4: Factor out $\frac{dy}{dx}$: $2 = \frac{dy}{dx}(\cos y - 3)$ Final answer: $\frac{dy}{dx} = \frac{2}{\cos y - 3}$
• Q: Find $\frac{dy}{dx}$ for $ax + by^2 = \cos(y)$. A: Step 1: Differentiate both sides with respect to $x$, treating $a$ and $b$ as constants: $\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)$ Step 2: Apply chain rule on $y$-dependent terms: $a + 2by\frac{dy}{dx} = -\sin(y)\frac{dy}{dx}$ Step 3: Move all terms with $\frac{dy}{dx}$ to the left-hand side: $2by\frac{dy}{dx} + \sin(y)\frac{dy}{dx} = -a$ Step 4: Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2by + \sin y) = -a$ Final answer: $\frac{dy}{dx} = -\frac{a}{2by + \sin y}$
• Q: Find $\frac{dy}{dx}$ if $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1$. A: Step 1: Substitute $x = \tan\theta$, which implies $\theta = \tan^{-1}x$. Under the given interval $0 < x < 1$, $\theta$ lies in the interval $(0, \pi/4)$. Step 2: Substitute $x$ into the expression: $y = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)$ Step 3: Apply double-angle formula $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$: $y = \cos^{-1}(\cos 2\theta) = 2\theta$ (since $0 < 2\theta < \pi/2$, which is inside the principal value branch of cosine) Step 4: Replace $\theta$ with $\tan^{-1}x$: $y = 2\tan^{-1}x$ Step 5: Differentiate with respect to $x$: Final answer: $\frac{dy}{dx} = \frac{2}{1+x^2}$
• Q: Find $\frac{dy}{dx}$ if $y = \sec^{-1}\left(\frac{1}{2x^2-1}\right), 0 < x < \frac{1}{\sqrt{2}}$. A: Step 1: Substitute $x = \cos\theta$, which implies $\theta = \cos^{-1}x$. For $0 < x < \frac{1}{\sqrt{2}}$, we have $\frac{\pi}{4} < \theta < \frac{\pi}{2}$. Step 2: Substitute into the equation: $y = \sec^{-1}\left(\frac{1}{2\cos^2\theta - 1}\right)$ Step 3: Use the identity $2\cos^2\theta - 1 = \cos(2\theta)$: $y = \sec^{-1}\left(\frac{1}{\cos 2\theta}\right) = \sec^{-1}(\sec 2\theta)$ Step 4: Since $\frac{\pi}{2} < 2\theta < \pi$, which is within the range of principal values of $\sec^{-1}$, we write: $y = 2\theta = 2\cos^{-1}x$ Step 5: Differentiate with respect to $x$: $\frac{dy}{dx} = 2 \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$ Final answer: $\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}$

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