Question

If $\tan^{-1}(x^2 + y^2) = a^2$, then find $\frac{dy}{dx}$.

Hint:

When differentiating inverse trigonometric functions, remember to apply the chain rule and carefully differentiate the inside functions.

Answer 1

Differentiating the equation $\tan^{-1}(x^2 + y^2) = a^2$ with respect to $x$: \[\frac{d}{dx} \left[ \tan^{-1}(x^2 + y^2) \right] = \frac{d}{dx} [a^2]\]Applying the chain rule to the left side yields:\[\frac{1}{1 + (x^2 + y^2)^2} \cdot \frac{d}{dx}(x^2 + y^2) = 0\]Substituting $\frac{d}{dx}(x^2 + y^2) = 2x + 2y\frac{dy}{dx}$ and solving for $\frac{dy}{dx}$:\[\frac{1}{1 + (x^2 + y^2)^2} \cdot (2x + 2y \frac{dy}{dx}) = 0\]The solution for $\frac{dy}{dx}$ is:\[\frac{dy}{dx} = \frac{-2x}{2y}\]Therefore, $\frac{dy}{dx} = \frac{-x}{y}$