Question

for each problem, use implicit differentiation to find \\(\frac{dy}{dx}\\) in terms of \\(x\\) and \\(y\\).

1. \\(2x^3 = 2y^2 + 5\\)
2. \\(3x^2 + 3y^2 = 2\\)
3. \\(5y^2 = 2x^3 - 5y\\)
4. \\(4x^2 = 2y^3 + 4y\\)
5. \\(5x^3 = -3xy + 2\\)
6. \\(1 = 3x + 2x^2y^2\\)

Explanation:

Differentiate problem 1

We differentiate both sides of \(2x^3 = 2y^2 + 5\) with respect to \(x\).
Using the power rule and chain rule:

$$ 6x^2 = 4y \frac{dy}{dx} $$

Solving for \(\frac{dy}{dx}\):

$$ \frac{dy}{dx} = \frac{6x^2}{4y} = \frac{3x^2}{2y} $$

Differentiate problem 2

We differentiate both sides of \(3x^2 + 3y^2 = 2\) with respect to \(x\).
Using the power rule and chain rule:

$$ 6x + 6y \frac{dy}{dx} = 0 $$

Solving for \(\frac{dy}{dx}\):

$$ 6y \frac{dy}{dx} = -6x \implies \frac{dy}{dx} = -\frac{x}{y} $$

Differentiate problem 3

We differentiate both sides of \(5y^2 = 2x^3 - 5y\) with respect to \(x\).
Using the power rule and chain rule:

$$ 10y \frac{dy}{dx} = 6x^2 - 5 \frac{dy}{dx} $$

Grouping the \(\frac{dy}{dx}\) terms:

$$ (10y + 5) \frac{dy}{dx} = 6x^2 \implies \frac{dy}{dx} = \frac{6x^2}{10y + 5} $$

Differentiate problem 4

We differentiate both sides of \(4x^2 = 2y^3 + 4y\) with respect to \(x\).
Using the power rule and chain rule:

$$ 8x = 6y^2 \frac{dy}{dx} + 4 \frac{dy}{dx} $$

Factoring out \(\frac{dy}{dx}\):

$$ 8x = (6y^2 + 4) \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{8x}{6y^2 + 4} = \frac{4x}{3y^2 + 2} $$

Differentiate problems 5 and 6

For problem 5, \(5x^3 = -3xy + 2\). Differentiating with respect to \(x\) using the product rule:

$$ 15x^2 = -3y - 3x \frac{dy}{dx} \implies 3x \frac{dy}{dx} = -3y - 15x^2 \implies \frac{dy}{dx} = \frac{-y - 5x^2}{x} $$

For problem 6, \(1 = 3x + 2x^2y^2\). Differentiating with respect to \(x\) using the product rule:

$$ 0 = 3 + 4xy^2 + 4x^2y \frac{dy}{dx} \implies 4x^2y \frac{dy}{dx} = -3 - 4xy^2 \implies \frac{dy}{dx} = \frac{-3 - 4xy^2}{4x^2y} $$

Answer:

Question 1

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

Question 2

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

Question 3

$$\frac{dy}{dx} = \frac{6x^2}{10y + 5}$$

Question 4

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

Question 5

$$\frac{dy}{dx} = \frac{-5x^2 - y}{x}$$

Question 6

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