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Let
y=sqrtxe^(x).

(dy)/(dx)=

Let $y=\sqrt{x} e^{x}$ . $\newline$ $\frac{d y}{d x}=$

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Csc, sec, and cot of special angles

Full solution

Q. Let

y=\sqrt{x} e^{x}

.

\newline

\frac{d y}{d x}=

Identify Product Rule: To find the derivative of $y$ with respect to $x$ , we need to use the product rule because $y$ is a product of two functions, $\sqrt{x}$ and $e^{x}$ .
Define Functions: Let $u = \sqrt{x}$ and $v = e^{x}$ . Then, we need to find the derivatives of $u$ and $v$ with respect to $x$ .
Find Derivatives: The derivative of $u$ with respect to $x$ is $(1/2)x^{(-1/2)}$ because the derivative of $x^{(1/2)}$ is $(1/2)x^{(1/2 - 1)}$ .
Apply Product Rule: The derivative of $v$ with respect to $x$ is $e^{x}$ because the derivative of $e^{x}$ with respect to $x$ is $e^{x}$ .
Substitute into Formula: Now, apply the product rule: $(\frac{dy}{dx}) = u'(x)v(x) + u(x)v'(x)$ .
Simplify Expression: Substitute the derivatives and functions into the product rule: $(\frac{dy}{dx}) = (\frac{1}{2})x^{(-\frac{1}{2})}e^{x} + \sqrt{x}e^{x}$ .
Combine Terms: Simplify the expression: $(\frac{dy}{dx}) = (\frac{1}{2})x^{(-\frac{1}{2})}e^{(x)} + x^{(\frac{1}{2})}e^{(x)}.$
Final Answer: Combine the terms: $\frac{dy}{dx} = e^{x}\left(\frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}}\right)$ .
Final Answer: Combine the terms: $(dy)/(dx) = e^{x}((1/2)x^{(-1/2)} + x^{(1/2)})$ .The final answer is $(dy)/(dx) = e^{x}((1/2)x^{(-1/2)} + x^{(1/2)})$ .

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