Math / Calculus

QuestionFind $y$ " by implicit differentiation. Simplify where possible. $y^{\prime \prime}=\square$

Studdy Solution

STEP 1

1. The equation given is $x^2 + 5y^2 = 5$ .
2. We are tasked with finding the second derivative $y''$ using implicit differentiation.

STEP 2

1. Differentiate the given equation implicitly with respect to $x$ to find $y'$ .
2. Differentiate the expression for $y'$ implicitly to find $y''$ .
3. Simplify the expression for $y''$ .

STEP 3

Differentiate the given equation $x^2 + 5y^2 = 5$ implicitly with respect to $x$ :
$\frac{d}{dx}(x^2) + \frac{d}{dx}(5y^2) = \frac{d}{dx}(5)$
This becomes:
$2x + 10y \frac{dy}{dx} = 0$
Solve for $\frac{dy}{dx}$ :
$10y \frac{dy}{dx} = -2x$
$\frac{dy}{dx} = -\frac{2x}{10y} = -\frac{x}{5y}$

STEP 4

Differentiate $\frac{dy}{dx} = -\frac{x}{5y}$ implicitly with respect to $x$ to find $y''$ :
Apply the quotient rule:
$\frac{d}{dx}\left(-\frac{x}{5y}\right) = -\frac{(5y)(1) - (x)(5\frac{dy}{dx})}{(5y)^2}$
Substitute $\frac{dy}{dx} = -\frac{x}{5y}$ :
$y'' = -\frac{5y - x \left(5\left(-\frac{x}{5y}\right)\right)}{25y^2}$
Simplify the expression:
$y'' = -\frac{5y + x\left(\frac{5x}{5y}\right)}{25y^2}$
$y'' = -\frac{5y + \frac{x^2}{y}}{25y^2}$

STEP 5

Simplify the expression for $y''$ :
Combine terms in the numerator:
$y'' = -\frac{5y^2 + x^2}{25y^3}$
The second derivative $y''$ is:
$y'' = -\frac{5y^2 + x^2}{25y^3}$

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