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We are tasked with finding the second derivative \( \frac{d^2 y}{dx^2} \) in terms of \( x \) and \( y \), given the equation:

\[
y^7 = x^8
\]

### Step 1: Differentiate the given equation implicitly
First, we differentiate both sides of the equation with respect to \( x \) using implicit differentiation.

\[
\frac{d}{dx} \left( y^7 \right) = \frac{d}{dx} \left( x^8 \right)
\]

Using the chain rule, we get:

\[
7y^6 \frac{dy}{dx} = 8x^7
\]

Now, solve for \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{8x^7}{7y^6}
\]

### Step 2: Differentiate again to find \( \frac{d^2 y}{dx^2} \)
Now, we take the derivative of \( \frac{dy}{dx} \) to find \( \frac{d^2 y}{dx^2} \).

\[
\frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{8x^7}{7y^6} \right)
\]

Using the quotient rule, we differentiate:

\[
\frac{d}{dx} \left( \frac{8x^7}{7y^6} \right) = \frac{ \left( 7y^6 \cdot 8 \cdot 7x^6 \right) - \left( 8x^7 \cdot 6y^5 \cdot \frac{dy}{dx} \right) }{(7y^6)^2}
\]

Substitute \( \frac{dy}{dx} = \frac{8x^7}{7y^6} \) from the previous step, simplify the expression, and you'll have the second derivative \( \frac{d^2 y}{dx^2} \) in terms of \( x \) and \( y \).

Would you like further clarification on any steps or a deeper explanation?

Here are 5 related questions to expand your understanding:
1. How does implicit differentiation work for higher-order derivatives?
2. What is the quotient rule in differentiation?
3. Can you explain the chain rule in more detail?
4. How would you apply this method to different implicit functions?
5. How does the power rule simplify derivatives in equations involving exponents?

**Tip**: Always keep track of derivatives of both \( y \) and \( x \) while performing implicit differentiation.

Find the second derivative of y in terms of x and y, given y^7 = x^8.

Learn how to find the second derivative of y in terms of x and y given the equation y^7 = x^8. This problem involves implicit differentiation, the chain rule, and the quotient rule, providing a step-by-step solution. Suitable for students in Grades 11-12.

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