To differentiate the function \( F(x) = \left( \frac{1}{x} + 1 \right)(x - 1) \) using the product rule, we will follow these steps:
Step 1: Identify the two functions for the product rule
Let:
- \( u = \frac{1}{x} + 1 \)
- \( v = x - 1 \)
Step 2: Differentiate \( u \) and \( v \)
Differentiating \( u \):
\[ u = \frac{1}{x} + 1 \] To differentiate \( u \), we find the derivative of \( \frac{1}{x} \): \[ \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \] So, \[ \frac{du}{dx} = -\frac{1}{x^2} \]
Differentiating \( v \):
\[ v = x - 1 \] The derivative of \( v \) is straightforward: \[ \frac{dv}{dx} = 1 \]
Step 3: Apply the product rule
The product rule states that: \[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ F'(x) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \]
Step 4: Substitute the derivatives into the product rule formula
Substituting the known values:
- \( u = \frac{1}{x} + 1 \)
- \( \frac{dv}{dx} = 1 \)
- \( v = x - 1 \)
- \( \frac{du}{dx} = -\frac{1}{x^2} \)
This gives us:
\[ F'(x) = \left(\frac{1}{x} + 1\right)(1) + (x - 1)\left(-\frac{1}{x^2}\right) \]
Step 5: Simplifying the expression
Now simplifying the expression: \[ F'(x) = \frac{1}{x} + 1 - \frac{x - 1}{x^2} \]
Now let's focus on simplifying \(-\frac{x - 1}{x^2}\): \[ -\frac{x - 1}{x^2} = -\frac{x}{x^2} + \frac{1}{x^2} = -\frac{1}{x} + \frac{1}{x^2} \]
Step 6: Combine all parts
Now, combining everything together: \[ F'(x) = \frac{1}{x} + 1 - \frac{1}{x} + \frac{1}{x^2} \]
The \(\frac{1}{x}\) cancels out: \[ F'(x) = 1 + \frac{1}{x^2} \]
Final result
Thus, the derivative of \( F(x) \) is: \[ F'(x) = 1 + \frac{1}{x^2} \]