If ` y^(2) = ax^(2) + b , " then " (d^(2)y)/( dx^(2))`

Options

A. ` (ab)/x^(3) `
B. ` x^(3)/(ab)`
C. ` (ab)/y^(2)`
D. ` (ab)/(y^(3)`

Answer

To solve the problem \( y^2 = ax^2 + b \) and find \( \frac{d^2y}{dx^2} \), we can follow these steps:

### Step 1: Differentiate \( y^2 = ax^2 + b \)

We start by differentiating both sides with respect to \( x \):

\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(ax^2 + b) \]

Using the chain rule on the left side and the power rule on the right side, we get:

\[ 2y \frac{dy}{dx} = 2ax \]

### Step 2: Solve for \( \frac{dy}{dx} \)

Now, we can isolate \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{2ax}{2y} = \frac{ax}{y} \]

### Step 3: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

Next, we differentiate \( \frac{dy}{dx} = \frac{ax}{y} \) with respect to \( x \):

Using the quotient rule, where \( u = ax \) and \( v = y \):

\[ \frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

\[ \frac{du}{dx} = a \]
\[ \frac{dv}{dx} = \frac{dy}{dx} = \frac{ax}{y} \]

Substituting these into the quotient rule:

\[ \frac{d^2y}{dx^2} = \frac{y(a) - ax\left(\frac{ax}{y}\right)}{y^2} \]

### Step 4: Simplify the expression

Now simplifying the numerator:

\[ = \frac{ay - \frac{a^2x^2}{y}}{y^2} \]

Multiplying through by \( y \):

\[ = \frac{ay^2 - a^2x^2}{y^3} \]

### Step 5: Substitute \( y^2 \)

Recall that \( y^2 = ax^2 + b \). Substitute this into the expression:

\[ = \frac{a(ax^2 + b) - a^2x^2}{y^3} \]

### Step 6: Final simplification

This simplifies to:

\[ = \frac{a^2x^2 + ab - a^2x^2}{y^3} = \frac{ab}{y^3} \]

Thus, we have:

\[ \frac{d^2y}{dx^2} = \frac{ab}{y^3} \]

### Conclusion

The final result is:

\[ \frac{d^2y}{dx^2} = \frac{ab}{y^3} \]