Let f(x)=x^(2)e^(x). f^(')(x)=

Identify rule: Identify the rule to use for differentiation; here, we need the product rule since f(x) is the product of two functions, x^2 and e^x.Apply product rule: Apply the product rule: (u*v)' = u'v + uv', where u = x^2 and v = e^x.Differentiate u: Differentiate u = x^2 to get u' = 2x.Differentiate v: Differentiate v = e^x to get v' = e^x.Plug into formula: Plug u', u, v', and v into the product rule formula: f'(x) = 2x * e^x + x^2 * e^x.Simplify expression: Combine like terms to simplify the expression: f'(x) = (2x + x^2) * e^x.Correct mistake: Realize there's a mistake in the previous step; we don't actually combine terms like that. Correct the expression: f'(x) = 2x * e^x + x^2 * e^x.

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Let $f(x)=x^{2} e^{x}$ . $\newline$ $f^{\prime}(x)=$

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Algebra 2

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Q. Let

f(x)=x^{2} e^{x}

\newline

f^{\prime}(x)=

Identify rule: Identify the rule to use for differentiation; here, we need the product rule since $f(x)$ is the product of two functions, $x^2$ and $e^x$ .
Apply product rule: Apply the product rule: $(u*v)' = u'v + uv'$ , where $u = x^2$ and $v = e^x$ .
Differentiate $u$ : Differentiate $u = x^2$ to get $u' = 2x$ .
Differentiate $v$ : Differentiate $v = e^x$ to get $v' = e^x$ .
Plug into formula: Plug $u'$ , $u$ , $v'$ , and $v$ into the product rule formula: $f'(x) = 2x \cdot e^x + x^2 \cdot e^x$ .
Simplify expression: Combine like terms to simplify the expression: $f'(x) = (2x + x^2) \cdot e^x$ .
Correct mistake: Realize there's a mistake in the previous step; we don't actually combine terms like that. Correct the expression: $f'(x) = 2x \cdot e^x + x^2 \cdot e^x$ .