Math / Calculus

Question $1 = 3x + 2x^2y^2$ Implicit differentiation

Studdy Solution

STEP 1

1. We are given an implicit function $1 = 3x + 2x^2y^2$ .
2. We need to find the derivative $\frac{dy}{dx}$ using implicit differentiation.
3. Both $x$ and $y$ are functions of some variable, typically $x$ .

STEP 2

1. Differentiate both sides of the equation with respect to $x$ .
2. Apply the product rule to the term $2x^2y^2$ .
3. Solve for $\frac{dy}{dx}$ .

STEP 3

Differentiate both sides of the equation with respect to $x$ .
The left side of the equation is a constant, so its derivative is zero:
$\frac{d}{dx}(1) = 0$
Differentiate the right side:
$\frac{d}{dx}(3x + 2x^2y^2)$

STEP 4

Differentiate $3x$ with respect to $x$ :
$\frac{d}{dx}(3x) = 3$

STEP 5

Differentiate $2x^2y^2$ using the product rule. The product rule states that $\frac{d}{dx}(uv) = u'v + uv'$ .
Let $u = 2x^2$ and $v = y^2$ .
First, find the derivative of $u$ with respect to $x$ :
$u = 2x^2$ $u' = \frac{d}{dx}(2x^2) = 4x$
Next, find the derivative of $v$ with respect to $x$ using the chain rule:
$v = y^2$ $v' = \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$
Apply the product rule:
$\frac{d}{dx}(2x^2y^2) = 4x \cdot y^2 + 2x^2 \cdot 2y \cdot \frac{dy}{dx}$
Simplify:
$= 4xy^2 + 4x^2y \cdot \frac{dy}{dx}$

STEP 6

Combine the derivatives from the previous steps:
$0 = 3 + 4xy^2 + 4x^2y \cdot \frac{dy}{dx}$
Solve for $\frac{dy}{dx}$ :
$0 - 3 - 4xy^2 = 4x^2y \cdot \frac{dy}{dx}$
$-3 - 4xy^2 = 4x^2y \cdot \frac{dy}{dx}$
Divide both sides by $4x^2y$ to isolate $\frac{dy}{dx}$ :
$\frac{dy}{dx} = \frac{-3 - 4xy^2}{4x^2y}$
The derivative $\frac{dy}{dx}$ is:
$\frac{dy}{dx} = \frac{-3 - 4xy^2}{4x^2y}$

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