f is a differentiable function such that `x=f(t^2),y=f(t^3)` and `f'(1)!=0` if `((d^2y)/(dx^2))_(t=1)`=

Options

A. `3/4 ((f''(1)+f (1))/((f' (1))^(2)))`
B. `3/4 ((f (1). f '(1) -f'(1))/(f (f'(1))^(2)))`
C. `4/3 (f'(1))/((f'(1))^(2))`
D. `4/3((f '(1)f''(1) -f' (1))/((f'(1))^(2)))`

Answer

To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \) at \( t = 1 \) given that \( x = f(t^2) \) and \( y = f(t^3) \). We will use the chain rule and implicit differentiation.

### Step-by-Step Solution:

1. **Differentiate \( x \) with respect to \( t \)**:
\[ x = f(t^2) \implies \frac{dx}{dt} = f'(t^2) \cdot \frac{d(t^2)}{dt} = f'(t^2) \cdot 2t \]
- **Hint**: Use the chain rule for differentiation.

2. **Differentiate \( y \) with respect to \( t \)**:
\[ y = f(t^3) \implies \frac{dy}{dt} = f'(t^3) \cdot \frac{d(t^3)}{dt} = f'(t^3) \cdot 3t^2 \]
- **Hint**: Again, apply the chain rule for differentiation.

3. **Find \( \frac{dy}{dx} \)**:
Using the relationship \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \):
\[ \frac{dy}{dx} = \frac{f'(t^3) \cdot 3t^2}{f'(t^2) \cdot 2t} = \frac{3t f'(t^3)}{2f'(t^2)} \]
- **Hint**: Remember to simplify the fractions carefully.

4. **Differentiate \( \frac{dy}{dx} \) with respect to \( t \)**:
We will use the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dx}{dt}} \]
Let \( u = 3t f'(t^3) \) and \( v = 2f'(t^2) \):
\[ \frac{d^2y}{dx^2} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \cdot \frac{1}{\frac{dx}{dt}} \]
- **Hint**: Apply the product rule and quotient rule carefully.

5. **Evaluate \( \frac{d^2y}{dx^2} \) at \( t = 1 \)**:
Substitute \( t = 1 \):
\[ \frac{d^2y}{dx^2}\bigg|_{t=1} = \frac{2f'(1) \left(3f'(1) + 3f''(1)\right) - 3f'(1) \cdot 2f''(1)}{(2f'(1))^2} \]
Simplifying this expression gives:
\[ = \frac{3f'(1) + 3f''(1) - 2f''(1)}{2f'(1)} = \frac{3f'(1) + f''(1)}{2f'(1)} \]
- **Hint**: Substitute and simplify carefully.

6. **Final Result**:
The value of \( \frac{d^2y}{dx^2} \) at \( t = 1 \) is:
\[ \frac{3f'(1) + f''(1)}{2f'(1)} \]

### Final Answer:
\[ \frac{3f'(1) + f''(1)}{2f'(1)} \]