The problem asks us to find the derivatives of two functions. i. $3x + y^3 - 4y = 10x^2$ ii. $y = \sqrt[3]{3x-1}$

1. Problem Description

The problem asks us to find the derivatives of two functions.

i. $3x + y^3 - 4y = 10x^2$

ii. $y = \sqrt[3]{3x-1}$

2. Solution Steps

i. We will use implicit differentiation to find $\frac{dy}{dx}$.

Differentiate both sides of the equation $3x + y^3 - 4y = 10x^2$ with respect to $x$.

$\frac{d}{dx}(3x) + \frac{d}{dx}(y^3) - \frac{d}{dx}(4y) = \frac{d}{dx}(10x^2)$

Using the chain rule, we get:

$3 + 3y^2 \frac{dy}{dx} - 4\frac{dy}{dx} = 20x$

Now, isolate $\frac{dy}{dx}$:

$3y^2 \frac{dy}{dx} - 4\frac{dy}{dx} = 20x - 3$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx}(3y^2 - 4) = 20x - 3$

Finally, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{20x - 3}{3y^2 - 4}$

ii. We want to find the derivative of $y = \sqrt[3]{3x-1}$.

We can rewrite this as $y = (3x-1)^{\frac{1}{3}}$.

Using the chain rule, we have:

$\frac{dy}{dx} = \frac{1}{3}(3x-1)^{\frac{1}{3} - 1} \cdot \frac{d}{dx}(3x-1)$

$\frac{dy}{dx} = \frac{1}{3}(3x-1)^{-\frac{2}{3}} \cdot (3)$

$\frac{dy}{dx} = (3x-1)^{-\frac{2}{3}}$

$\frac{dy}{dx} = \frac{1}{(3x-1)^{\frac{2}{3}}}$

$\frac{dy}{dx} = \frac{1}{(\sqrt[3]{3x-1})^2}$

3. Final Answer

i. $\frac{dy}{dx} = \frac{20x - 3}{3y^2 - 4}$

ii. $\frac{dy}{dx} = \frac{1}{(3x-1)^{\frac{2}{3}}}$ or $\frac{1}{(\sqrt[3]{3x-1})^2}$