Let's work through the derivatives one by one from the image you provided. Here's the solution for each function:

1. $y = 5x^3 - 4x^2 + 3x - 6$

This is a polynomial function, and we can differentiate it term by term.

$\frac{dy}{dx} = \frac{d}{dx}(5x^3) - \frac{d}{dx}(4x^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(6)$

Applying the power rule:

$\frac{dy}{dx} = 15x^2 - 8x + 3$

2. $y = \sqrt[3]{2x} - 7$

First, rewrite $\sqrt[3]{2x}$ as $(2x)^{1/3}$. Now apply the chain rule.

$\frac{dy}{dx} = \frac{d}{dx} \left( (2x)^{1/3} \right)$

Using the chain rule and the power rule:

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

Thus,

[ \frac{dy}{dx} = \frac{2}{3\sqrt[3]{(2x)^2}} = \frac{2}{3\sqrt[3]{4x^2}} ]

3. $y = \frac{3x+1}{\sqrt{3x^2+2}}$

We will apply the quotient rule here:

The quotient rule states: $\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$

Let:

• $f(x) = 3x+1$, so $f'(x) = 3$
• $g(x) = \sqrt{3x^2+2} = (3x^2+2)^{1/2}$, so $g'(x) = \frac{1}{2}(3x^2+2)^{-1/2} \cdot 6x = \frac{6x}{2\sqrt{3x^2+2}} = \frac{3x}{\sqrt{3x^2+2}}$

Now apply the quotient rule:

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

Simplify:

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

This can be further simplified, but this is the main result.

Would you like to proceed with the remaining problems or need further details for the above? Here are some related problems for practice:

1. Differentiate $y = x^4 + 2x^3 - x + 5$.
2. Use the chain rule to differentiate $y = \sqrt{4x^2 + 1}$.
3. Find the derivative of $y = \frac{x^2 + 2}{x + 1}$.
4. Apply the product rule to differentiate $y = x^2 \cdot \ln(x)$.
5. Use implicit differentiation for $y^2 + x^2 = 1$.

Tip:

For functions involving products or quotients, always consider whether you need the product rule, quotient rule, or chain rule before proceeding.