Let the function y = f(x) be given by
` x= t^(5) -5t^(3) -20t +7`
` and y= 4t^(3) -3t^(2) -18t + 3`
where ` t in ( -2,2) ` then f'(x) at t = 1 is

Options

A. ` 5/2 `
B. `2/5`
C. `7/5`
D. `5/7`

Answer

To find \( f'(x) \) at \( t = 1 \) for the given parametric equations, we will follow these steps:

1. **Identify the given equations**:
We have:
\[ x = t^5 - 5t^3 - 20t + 7 \]
\[ y = 4t^3 - 3t^2 - 18t + 3 \]

2. **Differentiate \( x \) with respect to \( t \)**:
We need to find \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = \frac{d}{dt}(t^5 - 5t^3 - 20t + 7) \]
Using the power rule:
\[ \frac{dx}{dt} = 5t^4 - 15t^2 - 20 \]

3. **Evaluate \( \frac{dx}{dt} \) at \( t = 1 \)**:
Substitute \( t = 1 \):
\[ \frac{dx}{dt} \bigg|_{t=1} = 5(1)^4 - 15(1)^2 - 20 = 5 - 15 - 20 = -30 \]

4. **Differentiate \( y \) with respect to \( t \)**:
Now, we find \( \frac{dy}{dt} \):
\[ \frac{dy}{dt} = \frac{d}{dt}(4t^3 - 3t^2 - 18t + 3) \]
Again using the power rule:
\[ \frac{dy}{dt} = 12t^2 - 6t - 18 \]

5. **Evaluate \( \frac{dy}{dt} \) at \( t = 1 \)**:
Substitute \( t = 1 \):
\[ \frac{dy}{dt} \bigg|_{t=1} = 12(1)^2 - 6(1) - 18 = 12 - 6 - 18 = -12 \]

6. **Find \( \frac{dy}{dx} \)**:
We can find \( \frac{dy}{dx} \) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substitute the values we found:
\[ \frac{dy}{dx} = \frac{-12}{-30} = \frac{12}{30} = \frac{2}{5} \]

7. **Conclusion**:
Therefore, \( f'(x) \) at \( t = 1 \) is:
\[ f'(x) = \frac{2}{5} \]