Let $f(x) = \frac{e^x}{x^2}$. We'll calculate $f'(x)$ (first derivative) and $f''(x)$ (second derivative), simplify, and factorize where possible.

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Step 1: First Derivative $f'(x)$

Using the quotient rule, $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$, where:

• $u = e^x$, $u' = e^x$,
• $v = x^2$, $v' = 2x$,

we get: $f'(x) = \frac{e^x(x^2) - e^x(2x)}{x^4}$

Simplify the numerator: $f'(x) = \frac{e^x(x^2 - 2x)}{x^4}$

Factorize the numerator: $f'(x) = \frac{e^x x(x - 2)}{x^4}$

Simplify: $f'(x) = \frac{e^x(x - 2)}{x^3}$

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Step 2: Second Derivative $f''(x)$

Now differentiate $f'(x) = \frac{e^x(x - 2)}{x^3}$ again, using the quotient rule:

• $u = e^x(x - 2)$, $u' = e^x(x - 2) + e^x \cdot 1 = e^x(x - 1)$,
• $v = x^3$, $v' = 3x^2$,

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

Substitute: $f''(x) = \frac{e^x(x - 1)x^3 - e^x(x - 2)3x^2}{x^6}$

Simplify the numerator: $\text{Numerator} = e^x(x - 1)x^3 - 3e^x(x - 2)x^2$

Factorize $e^x x^2$ from both terms: $\text{Numerator} = e^x x^2 \left[ x(x - 1) - 3(x - 2) \right]$

Simplify inside the brackets: $x(x - 1) - 3(x - 2) = x^2 - x - 3x + 6 = x^2 - 4x + 6$

Thus: $\text{Numerator} = e^x x^2 (x^2 - 4x + 6)$

Simplify $f''(x)$: $f''(x) = \frac{e^x x^2 (x^2 - 4x + 6)}{x^6}$

Simplify further: $f''(x) = \frac{e^x (x^2 - 4x + 6)}{x^4}$

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Final Results:

1. First derivative: $f'(x) = \frac{e^x(x - 2)}{x^3}$

2. Second derivative: $f''(x) = \frac{e^x(x^2 - 4x + 6)}{x^4}$

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Let me know if you have any questions! 😊 Here are 5 related practice problems:

1. Differentiate $f(x) = \frac{\ln(x)}{x^2}$.
2. Find $f'(x)$ and $f''(x)$ for $f(x) = \frac{\sin(x)}{x^3}$.
3. Simplify and factorize $\frac{d}{dx}\left(\frac{e^{2x}}{x^2}\right)$.
4. Calculate the second derivative of $g(x) = \frac{x^2}{e^x}$.
5. Solve for $f'(x)$ and $f''(x)$ for $f(x) = \frac{1}{x^n}$.

Tip: Always simplify the numerator step by step in the quotient rule; it avoids common errors!