Class CBSE Class 12 Mathematics Derivatives Q #1356
KNOWLEDGE BASED
APPLY
2 Marks 2025 AISSCE(Board Exam) VSA
If $\tan^{-1}(x^{2}+y^{2})=a^{2}$, then find $\frac{dy}{dx}$.
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Detailed Solution

Step 1: Differentiate both sides with respect to $x$

Given $\tan^{-1}(x^{2}+y^{2})=a^{2}$. Differentiating both sides with respect to $x$, we get: $$\frac{d}{dx} \left[ \tan^{-1}(x^{2}+y^{2}) \right] = \frac{d}{dx} (a^{2})$$

Step 2: Apply the chain rule

Using the chain rule, we have: $$\frac{1}{1+(x^{2}+y^{2})^{2}} \cdot \frac{d}{dx}(x^{2}+y^{2}) = 0$$ Since $a$ is a constant, its derivative is 0.

Step 3: Differentiate $x^2 + y^2$

Differentiating $x^{2}+y^{2}$ with respect to $x$, we get: $$\frac{d}{dx}(x^{2}+y^{2}) = 2x + 2y\frac{dy}{dx}$$

Step 4: Substitute back into the equation

Substituting this back into the equation, we have: $$\frac{1}{1+(x^{2}+y^{2})^{2}} \cdot \left(2x + 2y\frac{dy}{dx}\right) = 0$$

Step 5: Solve for $\frac{dy}{dx}$

Since $\frac{1}{1+(x^{2}+y^{2})^{2}}$ is never zero, we must have: $$2x + 2y\frac{dy}{dx} = 0$$ $$2y\frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = -\frac{2x}{2y}$$ $$\frac{dy}{dx} = -\frac{x}{y}$$

Final Answer: $-\frac{x}{y}$

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concepts of implicit differentiation and chain rule to find the derivative.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure of implicit differentiation and applying the chain rule to arrive at the solution.<\/span>
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of differentiation of inverse trigonometric functions and implicit functions, which are standard topics in the syllabus.

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