Math / Calculus

QuestionConsider the following equation. $6 x^{2}-y^{2}=4$ (a) Find $y^{\prime}$ by implicit differentiation. $y^{\prime}=$ $\square$ (b) Solve the equation explicitly for $y$ and differentiate to get $y^{\prime}$ in terms of $x$ . $y^{\prime}= \pm$ $\square$ Need Help? Read It Watch It Submit Answer 14. [-/4.76 Points] DETAILS MY NOTES SCALCET9 3.5.009.
Find $\frac{d y}{d x}$ by implicit differentiation. $\begin{array}{l} \frac{x^{2}}{x+y}=y^{2}+2 \\ \frac{d y}{d x}=\square \end{array}$

Studdy Solution

STEP 1

1. We are given two separate problems involving implicit differentiation.
2. For each problem, we will use implicit differentiation techniques to find $\frac{dy}{dx}$ .
3. We will solve the first problem in two parts: (a) implicit differentiation and (b) solving explicitly for $y$ .
4. The second problem involves finding $\frac{dy}{dx}$ directly using implicit differentiation.

STEP 2

1. Problem 1(a): Use implicit differentiation on $6x^2 - y^2 = 4$ .
2. Problem 1(b): Solve the equation explicitly for $y$ and differentiate.
3. Problem 2: Use implicit differentiation on $\frac{x^2}{x+y} = y^2 + 2$ .

STEP 3

Differentiate the equation $6x^2 - y^2 = 4$ implicitly with respect to $x$ :
- Differentiate $6x^2$ with respect to $x$ : $\frac{d}{dx}(6x^2) = 12x$ . - Differentiate $-y^2$ with respect to $x$ using the chain rule: $\frac{d}{dx}(-y^2) = -2y \cdot \frac{dy}{dx}$ . - Differentiate the constant $4$ : $\frac{d}{dx}(4) = 0$ .
Combine these to form the differentiated equation:
$12x - 2y \cdot \frac{dy}{dx} = 0$

STEP 4

Solve for $\frac{dy}{dx}$ :
$12x = 2y \cdot \frac{dy}{dx}$
Divide both sides by $2y$ :
$\frac{dy}{dx} = \frac{12x}{2y}$
Simplify:
$\frac{dy}{dx} = \frac{6x}{y}$

STEP 5

Solve the equation $6x^2 - y^2 = 4$ explicitly for $y$ :
$y^2 = 6x^2 - 4$
Take the square root of both sides:
$y = \pm \sqrt{6x^2 - 4}$
Differentiate $y$ with respect to $x$ :
For $y = \sqrt{6x^2 - 4}$ :
$\frac{dy}{dx} = \frac{1}{2\sqrt{6x^2 - 4}} \cdot (12x) = \frac{6x}{\sqrt{6x^2 - 4}}$
For $y = -\sqrt{6x^2 - 4}$ :
$\frac{dy}{dx} = -\frac{6x}{\sqrt{6x^2 - 4}}$
Thus, $y' = \pm \frac{6x}{\sqrt{6x^2 - 4}}$ .

STEP 6

Differentiate the equation $\frac{x^2}{x+y} = y^2 + 2$ implicitly with respect to $x$ :
- Differentiate $\frac{x^2}{x+y}$ using the quotient rule. - Differentiate $y^2$ using the chain rule.
Quotient rule for $\frac{x^2}{x+y}$ :
$\frac{d}{dx}\left(\frac{x^2}{x+y}\right) = \frac{(2x)(x+y) - x^2(1 + \frac{dy}{dx})}{(x+y)^2}$
Chain rule for $y^2$ :
$\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$
Combine these into the differentiated equation:
$\frac{2x(x+y) - x^2(1 + \frac{dy}{dx})}{(x+y)^2} = 2y \cdot \frac{dy}{dx}$

STEP 7

Solve for $\frac{dy}{dx}$ :
Rearrange and simplify to solve for $\frac{dy}{dx}$ :
$(2x(x+y) - x^2) - x^2 \frac{dy}{dx} = 2y(x+y)^2 \frac{dy}{dx}$
Combine terms involving $\frac{dy}{dx}$ and solve:
$\frac{dy}{dx} = \frac{2x(x+y) - x^2}{x^2 + 2y(x+y)^2}$
The solutions for $\frac{dy}{dx}$ are:
1. For Problem 1(a): $\frac{dy}{dx} = \frac{6x}{y}$
2. For Problem 1(b): $y' = \pm \frac{6x}{\sqrt{6x^2 - 4}}$
3. For Problem 2: $\frac{dy}{dx} = \frac{2x(x+y) - x^2}{x^2 + 2y(x+y)^2}$

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