Find the derivative of ln((xe^(2x))^0.5)

\frac{d}{dx}\left(\ln\left(\sqrt{x e^{2x}}\right)\right)

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Answer

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

Step by step solution

Problem

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

The power of a product is equal to the product of it's factors raised to the same power

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

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 $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{d}{dx}\left(\sqrt{x}e^x\right)\frac{1}{\sqrt{x}e^x}$
3

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=\sqrt{x}$ and $g=e^x$

$\left(\sqrt{x}\cdot\frac{d}{dx}\left(e^x\right)+e^x\frac{d}{dx}\left(\sqrt{x}\right)\right)\frac{1}{\sqrt{x}e^x}$
4

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\left(\sqrt{x}\cdot\frac{d}{dx}\left(e^x\right)+\frac{1}{2}x^{-\frac{1}{2}}e^x\right)\frac{1}{\sqrt{x}e^x}$
5

Applying the derivative of the exponential function

$\left(\sqrt{x}e^x+\frac{1}{2}x^{-\frac{1}{2}}e^x\right)\frac{1}{\sqrt{x}e^x}$
6

Multiplying the fraction and term

$\frac{\sqrt{x}e^x+\frac{1}{2}x^{-\frac{1}{2}}e^x}{\sqrt{x}e^x}$
7

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{\sqrt{x}e^x+\frac{1}{2}\cdot\frac{1}{\sqrt{x}}e^x}{\sqrt{x}e^x}$
8

Apply the formula: $a\frac{1}{x}$$=\frac{a}{x}$, where $a=\frac{1}{2}$ and $x=\sqrt{x}$

$\frac{\sqrt{x}e^x+\frac{\frac{1}{2}}{\sqrt{x}}e^x}{\sqrt{x}e^x}$
9

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

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

Answer

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

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.41 seconds

Views:

131