Q584: If $\tan^{-1}(x^{2}-y^{2})=a$, where 'a' is a constant, th...

Class CBSE Class 12 Mathematics Inverse Trigonometric Functions Q #584

COMPETENCY BASED

APPLY

1 Marks 2025 AISSCE(Board Exam) MCQ SINGLE

If $\tan^{-1}(x^{2}-y^{2})=a$, where 'a' is a constant, then $\frac{dy}{dx}$ is:

(A) $\frac{x}{y}$

(B) $-\frac{x}{y}$

(D) $\frac{a}{y}$

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AI Tutor Explanation

Detailed Solution

Step 1: Differentiate both sides with respect to x

Given $\tan^{-1}(x^{2}-y^{2})=a$, we differentiate both sides with respect to $x$.

Step 2: Apply the chain rule

Using the chain rule, we have: $$\frac{d}{dx} \tan^{-1}(x^{2}-y^{2}) = \frac{d}{dx} (a)$$ $$\frac{1}{1+(x^{2}-y^{2})^{2}} \cdot \frac{d}{dx}(x^{2}-y^{2}) = 0$$

Step 3: Differentiate the inner function

Now, differentiate $x^{2}-y^{2}$ with respect to $x$: $$\frac{d}{dx}(x^{2}-y^{2}) = 2x - 2y\frac{dy}{dx}$$

Step 4: Substitute back into the equation

Substitute this back into the equation: $$\frac{1}{1+(x^{2}-y^{2})^{2}} \cdot (2x - 2y\frac{dy}{dx}) = 0$$

Step 5: Solve for dy/dx

Since $\frac{1}{1+(x^{2}-y^{2})^{2}}$ cannot be 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: $\frac{x}{y}$

AI Suggestion: Option A

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because it requires the student to apply the concepts of implicit differentiation and trigonometric inverse functions 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.<\/span>

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It assesses the student's ability to apply differentiation techniques to solve a problem involving inverse trigonometric functions and implicit functions.