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 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, $\frac{dy}{dx} = \frac{-x}{y}$