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Step-by-step Solution

Find the derivative of $\ln\left(\sqrt{xe^{2x}}\right)$

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Answer

$\frac{\frac{1}{2}x^{-\frac{1}{2}}e^x+\sqrt{x}e^x}{\sqrt{x}}\cdot\frac{1}{e^x}$

Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\ln\left(\sqrt{x e^{2x}}\right)\right)$
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The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{\sqrt{x}\sqrt{e^{2x}}}\cdot\frac{d}{dx}\left(\sqrt{x}e^x\right)$

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Answer

$\frac{\frac{1}{2}x^{-\frac{1}{2}}e^x+\sqrt{x}e^x}{\sqrt{x}}\cdot\frac{1}{e^x}$
$\frac{d}{dx}\left(\ln\left(\sqrt{x e^{2x}}\right)\right)$

Main topic:

Differential calculus

Related formulas:

3. See formulas

Time to solve it:

~ 0.09 seconds

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