To find the derivative $f'(x)$ of the function $f(x) = \dfrac{x^2}{e^x}$, we will use the quotient rule. The quotient rule states that if you have a function of the form $\frac{u(x)}{v(x)}$, then the derivative is given by:

$\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

For the given function $f(x) = \frac{x^2}{e^x}$, we can identify:

• $u(x) = x^2$
• $v(x) = e^x$

Now, compute the derivatives of $u(x)$ and $v(x)$:

• $u'(x) = 2x$
• $v'(x) = e^x$

Using the quotient rule:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

Substituting the derivatives and functions:

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

Simplify the expression:

$f'(x) = \frac{2xe^x - x^2e^x}{e^{2x}}$

Factor out $e^x$ in the numerator:

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

Finally, cancel $e^x$ in the numerator and denominator:

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

So, the derivative is:

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

Would you like any further details or explanations?

Here are 5 related questions to expand on this topic:

1. How would you find the second derivative $f''(x)$ of the given function?
2. What is the significance of the quotient rule in calculus?
3. How would the derivative change if $f(x) = \dfrac{x^3}{e^x}$?
4. Can the derivative $f'(x)$ be simplified further, and what does that represent?
5. How does the behavior of $f(x)$ change as $x$ approaches infinity?

Tip: When applying the quotient rule, always simplify the expression at the end to make sure it’s in its most reduced form.