Question:
If $\tan^{-1}(x^2 + y^2) = a^2$, then find $\frac{dy}{dx}$.
If $\tan^{-1}(x^2 + y^2) = a^2$, then find $\frac{dy}{dx}$.
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When differentiating inverse trigonometric functions, remember to apply the chain rule and carefully differentiate the inside functions.
Updated On: Mar 6, 2026
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Solution and Explanation
Differentiating both sides of the equation $\tan^{-1}(x^2 + y^2) = a^2$ with respect to $x$, we get:
\[
\frac{d}{dx} \left[ \tan^{-1}(x^2 + y^2) \right] = \frac{d}{dx} [a^2]
\]
Using the chain rule on the left-hand side:
\[
\frac{1}{1 + (x^2 + y^2)^2} \cdot \frac{d}{dx}(x^2 + y^2) = 0
\]
Since $\frac{d}{dx}(x^2 + y^2) = 2x + 2y\frac{dy}{dx}$, we substitute this and solve for $\frac{dy}{dx}$:
\[
\frac{1}{1 + (x^2 + y^2)^2} \cdot (2x + 2y \frac{dy}{dx}) = 0
\]
Solving for $\frac{dy}{dx}$, we get:
\[
\frac{dy}{dx} = \frac{-2x}{2y}
\]
Thus, $\frac{dy}{dx} = \frac{-x}{y}$
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