To solve the problem and find the equation of the tangent line to the curve $y = f(x) = \frac{e^x}{x}$ at $x = 1$, follow these steps:

Step 1: Evaluate $f(1)$

The equation of the tangent line requires a point on the curve. Let's calculate $f(1)$:

$f(1) = \frac{e^1}{1} = e$

So, the point on the curve is $(1, e)$.

Step 2: Find the derivative $f'(x)$

The slope of the tangent line is given by the derivative $f'(x)$ evaluated at $x = 1$. Use the quotient rule to differentiate $f(x) = \frac{e^x}{x}$.

The quotient rule states: $\left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$ where $u(x) = e^x$ and $v(x) = x$.

• $u'(x) = e^x$
• $v'(x) = 1$

Now apply the quotient rule: $f'(x) = \frac{e^x \cdot x - e^x \cdot 1}{x^2} = \frac{e^x(x - 1)}{x^2}$

Step 3: Evaluate $f'(1)$

Substitute $x = 1$ into the derivative:

$f'(1) = \frac{e^1(1 - 1)}{1^2} = \frac{e(0)}{1} = 0$

So, the slope of the tangent line at $x = 1$ is 0.

Step 4: Equation of the tangent line

Since the slope of the tangent line is 0 and it passes through the point $(1, e)$, the equation of the tangent line is horizontal, i.e., $y = e$.

Final Answer:

The equation of the tangent line is $y = e$, which corresponds to option D.

Would you like more details or have any further questions?

Related Questions:

1. How is the quotient rule applied to differentiate functions?
2. What are the conditions for a tangent line to have a slope of zero?
3. Can you explain the general form of a tangent line equation?
4. How would the tangent line change if we considered another point on the curve?
5. What does the behavior of $\frac{e^x}{x}$ suggest about its asymptotes?

Tip:

The quotient rule is essential for differentiating functions of the form $\frac{f(x)}{g(x)}$. Always ensure that both $f(x)$ and $g(x)$ are properly differentiated before simplifying.