Finding the Second Derivative of y

To find the second derivative of \(y\) with respect to \(x\), denoted as \(y''\) or \(\frac{d^2y}{dx^2}\), given the equation \(x^4 + y^4 = 16\), we need to use the method of implicit differentiation twice.

First Derivative Calculation

We start by differentiating the given equation \(x^4 + y^4 = 16\) with respect to \(x\). Remember that when differentiating terms involving \(y\), we apply the chain rule, multiplying by \(\frac{dy}{dx}\) (or \(y'\)).

The equation is: \[x^4 + y^4 = 16\]

Differentiating both sides with respect to \(x\):

\[\frac{d}{dx}(x^4) + \frac{d}{dx}(y^4) = \frac{d}{dx}(16)\]

Applying the power rule and chain rule:

\[4x^3 + 4y^3 \frac{dy}{dx} = 0\]

Now, we solve for \(\frac{dy}{dx}\) (which is \(y'\)):

\[4y^3 \frac{dy}{dx} = -4x^3\]

\[\frac{dy}{dx} = \frac{-4x^3}{4y^3}\]

\[y' = \frac{-x^3}{y^3}\]

This is our first derivative.

Second Derivative Calculation

Next, we differentiate \(y' = \frac{-x^3}{y^3}\) with respect to \(x\) to find \(y''\). We will use the quotient rule, which states that for a function \(f(x) = \frac{u(x)}{v(x)}\), its derivative is \(f'(x) = \frac{u'v - uv'}{v^2}\).

Here, let \(u = -x^3\) and \(v = y^3\).

• Derivative of \(u\): \(u' = \frac{d}{dx}(-x^3) = -3x^2\)
• Derivative of \(v\): \(v' = \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} = 3y^2 y'\)

Now, apply the quotient rule to find \(y''\):

\[y'' = \frac{(-3x^2)(y^3) - (-x^3)(3y^2 y')}{(y^3)^2}\]

\[y'' = \frac{-3x^2 y^3 + 3x^3 y^2 y'}{y^6}\]

Substitute the expression for \(y'\) that we found earlier, \(y' = \frac{-x^3}{y^3}\):

\[y'' = \frac{-3x^2 y^3 + 3x^3 y^2 \left(\frac{-x^3}{y^3}\right)}{y^6}\]

Simplify the term inside the parenthesis:

\[y'' = \frac{-3x^2 y^3 + 3x^3 \left(\frac{-x^3}{y}\right)}{y^6}\]

\[y'' = \frac{-3x^2 y^3 - \frac{3x^6}{y}}{y^6}\]

To combine the terms in the numerator, find a common denominator (which is \(y\)):

\[y'' = \frac{\frac{-3x^2 y^3 \cdot y}{y} - \frac{3x^6}{y}}{y^6}\]

\[y'' = \frac{\frac{-3x^2 y^4 - 3x^6}{y}}{y^6}\]

Multiply the numerator by the reciprocal of the denominator:

\[y'' = \frac{-3x^2 y^4 - 3x^6}{y \cdot y^6}\]

\[y'' = \frac{-3x^2 y^4 - 3x^6}{y^7}\]

Factor out \(-3x^2\) from the numerator:

\[y'' = \frac{-3x^2 (y^4 + x^4)}{y^7}\]

Finally, recall the original equation given: \(x^4 + y^4 = 16\). Substitute this value into our expression for \(y''\):

\[y'' = \frac{-3x^2 (16)}{y^7}\]

\[y'' = \frac{-48x^2}{y^7}\]

Summary of Steps

The final calculated second derivative is \(y'' = \frac{-48x^2}{y^7}\).