Let \[f(x)=\dfrac{x^2}{e^x}\].

\[f'(x)=\] 




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.

Learn how to find the derivative of f(x) = x^2 / e^x using the quotient rule. This detailed step-by-step solution covers differentiation, the quotient rule, and exponential functions. Suitable for college-level mathematics.

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