The problem is to find the first derivative, $y'$, and the second derivative, $y''$, of $y$ with respect to $x$ for the equation $x^4 + y^4 = 16$.

1. Problem Description

The problem is to find the first derivative, $y'$, and the second derivative, $y''$, of $y$ with respect to $x$ for the equation $x^4 + y^4 = 16$.

2. Solution Steps

First, we find $y'$ using implicit differentiation. Differentiating both sides of the equation $x^4 + y^4 = 16$ with respect to $x$, we get:

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

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

Now, solve for $\frac{dy}{dx}$:

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

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

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

Next, we find $y''$ by differentiating $y'$ with respect to $x$:

$y'' = \frac{d}{dx} \left( -\frac{x^3}{y^3} \right) = -\frac{d}{dx} \left( \frac{x^3}{y^3} \right)$

Using the quotient rule, $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$, where $u = x^3$ and $v = y^3$, we have:

$y'' = - \frac{y^3(3x^2) - x^3(3y^2 \frac{dy}{dx})}{(y^3)^2}$

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

Substitute $y' = \frac{dy}{dx} = -\frac{x^3}{y^3}$ into the equation:

$y'' = - \frac{3x^2y^3 - 3x^3y^2 (-\frac{x^3}{y^3})}{y^6}$

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

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

$y'' = - \frac{\frac{3x^2y^4 + 3x^6}{y}}{y^6}$

$y'' = - \frac{3x^2y^4 + 3x^6}{y^7}$

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

Since $x^4 + y^4 = 16$, we substitute this into the equation for $y''$:

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

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

3. Final Answer

$y' = -\frac{x^3}{y^3}$

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