We are given the function:

$y = -\frac{e^x}{x}$

We need to find the derivative $y'$ with respect to $x$. To do this, we will apply the quotient rule for differentiation.

The quotient rule states that for a function of the form:

$y = \frac{u(x)}{v(x)}$

The derivative is:

$y' = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2}$

For our function, we have:

• $u(x) = -e^x$
• $v(x) = x$

Now, we differentiate each part:

• The derivative of $u(x) = -e^x$ is $u'(x) = -e^x$
• The derivative of $v(x) = x$ is $v'(x) = 1$

Applying the quotient rule:

$y' = \frac{x(-e^x) - (-e^x)(1)}{x^2}$

Simplifying the numerator:

$y' = \frac{-x e^x + e^x}{x^2}$

Factor out $e^x$ from the numerator:

$y' = \frac{e^x(-x + 1)}{x^2}$

Thus, the derivative is:

$y' = \frac{e^x(1 - x)}{x^2}$

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Do you want further details or have any questions about this solution?

Here are some related questions:

1. What happens to the function as $x$ approaches 0?
2. How do you find the second derivative of this function?
3. What is the behavior of $y'$ as $x$ becomes very large (positive or negative)?
4. Can the function $y = -\frac{e^x}{x}$ have any horizontal asymptotes?
5. How would you find the critical points of this function?

Tip: When applying the quotient rule, always differentiate the numerator and denominator separately before plugging them into the formula.