Question

If \( a(u + x^2) = x \) and \( y - x^3 = a^2 \), then \( \frac{dy}{dx} \) at \( x = 1 \) is:

Hint:

Use substitution and implicit differentiation where multiple variables are involved, and always simplify step by step before substituting values.

Answer 1

Given: \[ a(u + x^2) = x \quad \text{(1)} \] \[ y - x^3 = a^2 \quad \Rightarrow \quad y = a^2 + x^3 \quad \text{(2)} \] From (1), solve for \( a \): \[ a = \frac{x}{u + x^2} \] Now substitute \( a^2 \) into (2): \[ y = \left( \frac{x}{u + x^2} \right)^2 + x^3 \] Differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} \left( \left( \frac{x}{u + x^2} \right)^2 \right) + \frac{d}{dx}(x^3) \] Let \( f(x) = \left( \frac{x}{u + x^2} \right)^2 \). Using the chain rule: \[ \frac{d}{dx} f(x) = 2 \cdot \frac{x}{u + x^2} \cdot \frac{d}{dx} \left( \frac{x}{u + x^2} \right) \] Let: \[ \frac{d}{dx} \left( \frac{x}{u + x^2} \right) = \frac{(u + x^2)(1) - x(2x)}{(u + x^2)^2} = \frac{u + x^2 - 2x^2}{(u + x^2)^2} = \frac{u - x^2}{(u + x^2)^2} \] Now plug back into derivative: \[ \frac{dy}{dx} = 2 \cdot \frac{x}{u + x^2} \cdot \frac{u - x^2}{(u + x^2)^2} + 3x^2 \] At \( x = 1 \), from (1): \[ a(u + 1) = 1 \Rightarrow a = \frac{1}{u + 1} \Rightarrow a^2 = \frac{1}{(u + 1)^2} \] Substitute into final derivative expression to compute value at \( x = 1 \).