given the equation √(x⁶ + 9y²) = 3x + 2, find dy/dx by implicit differentiation.

Explanation:

Step1: Differentiate both sides

Differentiate $\sqrt{x^{6}+9y^{2}}$ and $3x + 2$ with respect to $x$. The derivative of the left - hand side using the chain rule: Let $u=x^{6}+9y^{2}$, so $\sqrt{u}=u^{\frac{1}{2}}$. The derivative of $\sqrt{u}$ with respect to $u$ is $\frac{1}{2}u^{-\frac{1}{2}}$, and the derivative of $u$ with respect to $x$ is $6x^{5}+18y\frac{dy}{dx}$. Then the derivative of $\sqrt{x^{6}+9y^{2}}$ with respect to $x$ is $\frac{1}{2\sqrt{x^{6}+9y^{2}}}(6x^{5}+18y\frac{dy}{dx})$. The derivative of the right - hand side $3x + 2$ with respect to $x$ is $3$. So we have $\frac{1}{2\sqrt{x^{6}+9y^{2}}}(6x^{5}+18y\frac{dy}{dx})=3$.

Step2: Multiply both sides by $2\sqrt{x^{6}+9y^{2}}$

To get rid of the denominator on the left - hand side. $6x^{5}+18y\frac{dy}{dx}=6\sqrt{x^{6}+9y^{2}}$.

Step3: Isolate $\frac{dy}{dx}$

Subtract $6x^{5}$ from both sides: $18y\frac{dy}{dx}=6\sqrt{x^{6}+9y^{2}}-6x^{5}$. Then divide both sides by $18y$ (assuming $y\neq0$). $\frac{dy}{dx}=\frac{6\sqrt{x^{6}+9y^{2}}-6x^{5}}{18y}=\frac{\sqrt{x^{6}+9y^{2}}-x^{5}}{3y}$.

Answer:

$\frac{\sqrt{x^{6}+9y^{2}}-x^{5}}{3y}$