`"Let "y=t^(10)+1 and x=t^(8)+1." Then "(d^(2)y)/(dx^(2))` is

Options

A. `(5)/(2)t`
B. `20t^(8)`
C. `(5)/(16t^(6))`
D. none of these

Explanation

To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \), given the equations \( y = t^{10} + 1 \) and \( x = t^{8} + 1 \).

### Step-by-Step Solution:

1. **Differentiate \( y \) with respect to \( t \)**:
\[ \frac{dy}{dt} = \frac{d}{dt}(t^{10} + 1) = 10t^{9} \]

2. **Differentiate \( x \) with respect to \( t \)**:
\[ \frac{dx}{dt} = \frac{d}{dt}(t^{8} + 1) = 8t^{7} \]

3. **Find \( \frac{dy}{dx} \)** using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{10t^{9}}{8t^{7}} = \frac{10}{8} t^{2} = \frac{5}{4} t^{2} \]

4. **Express \( t \) in terms of \( x \)**:
From the equation \( x = t^{8} + 1 \), we can express \( t^{8} \) as:
\[ t^{8} = x - 1 \]
Therefore,
\[ t = (x - 1)^{1/8} \]

5. **Substitute \( t \) back into \( \frac{dy}{dx} \)**:
\[ \frac{dy}{dx} = \frac{5}{4} \left((x - 1)^{1/8}\right)^{2} = \frac{5}{4} (x - 1)^{1/4} \]

6. **Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)**:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{5}{4} (x - 1)^{1/4}\right) \]
Using the power rule:
\[ \frac{d^2y}{dx^2} = \frac{5}{4} \cdot \frac{1}{4} (x - 1)^{-3/4} \cdot \frac{d}{dx}(x - 1) \]
Since \( \frac{d}{dx}(x - 1) = 1 \):
\[ \frac{d^2y}{dx^2} = \frac{5}{16} (x - 1)^{-3/4} \]

7. **Final expression**:
Therefore, the second derivative \( \frac{d^2y}{dx^2} \) is:
\[ \frac{d^2y}{dx^2} = \frac{5}{16 (x - 1)^{3/4}} \]