Find the derivative of ln(x^0.5(x-3)^0.5)

\frac{d}{dx}\left(\ln\left(\sqrt{x}\sqrt{x-3}\right)\right)

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Answer

$\frac{\sqrt{x}\cdot\frac{\frac{1}{2}}{\sqrt{x-3}}+\sqrt{x-3}\cdot\frac{\frac{1}{2}}{\sqrt{x}}}{\sqrt{x-3}\sqrt{x}}$

Step by step solution

Problem

$\frac{d}{dx}\left(\ln\left(\sqrt{x}\sqrt{x-3}\right)\right)$
1

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

$\frac{1}{\sqrt{x-3}\sqrt{x}}\cdot\frac{d}{dx}\left(\sqrt{x-3}\sqrt{x}\right)$
2

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=\sqrt{x}$ and $g=\sqrt{x-3}$

$\frac{1}{\sqrt{x-3}\sqrt{x}}\left(\sqrt{x}\cdot\frac{d}{dx}\left(\sqrt{x-3}\right)+\sqrt{x-3}\cdot\frac{d}{dx}\left(\sqrt{x}\right)\right)$
3

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

$\frac{1}{\sqrt{x-3}\sqrt{x}}\left(\sqrt{x}\cdot\frac{d}{dx}\left(\sqrt{x-3}\right)+\frac{1}{2}\sqrt{x-3}x^{-\frac{1}{2}}\right)$
4

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

$\frac{1}{\sqrt{x-3}\sqrt{x}}\left(\frac{1}{2}\sqrt{x}\left(x-3\right)^{-\frac{1}{2}}\cdot\frac{d}{dx}\left(x-3\right)+\frac{1}{2}\sqrt{x-3}x^{-\frac{1}{2}}\right)$
5

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{1}{\sqrt{x-3}\sqrt{x}}\left(\frac{1}{2}\sqrt{x}\left(x-3\right)^{-\frac{1}{2}}\left(\frac{d}{dx}\left(-3\right)+\frac{d}{dx}\left(x\right)\right)+\frac{1}{2}\sqrt{x-3}x^{-\frac{1}{2}}\right)$
6

The derivative of the constant function is equal to zero

$\frac{1}{\sqrt{x-3}\sqrt{x}}\left(\frac{1}{2}\sqrt{x}\left(x-3\right)^{-\frac{1}{2}}\left(0+\frac{d}{dx}\left(x\right)\right)+\frac{1}{2}\sqrt{x-3}x^{-\frac{1}{2}}\right)$
7

The derivative of the linear function is equal to $1$

$\left(\left(0+1\right)\cdot \frac{1}{2}\sqrt{x}\left(x-3\right)^{-\frac{1}{2}}+\frac{1}{2}\sqrt{x-3}x^{-\frac{1}{2}}\right)\frac{1}{\sqrt{x-3}\sqrt{x}}$
8

Add the values $1$ and $0$

$\left(1\cdot \frac{1}{2}\sqrt{x}\left(x-3\right)^{-\frac{1}{2}}+\frac{1}{2}\sqrt{x-3}x^{-\frac{1}{2}}\right)\frac{1}{\sqrt{x-3}\sqrt{x}}$
9

Multiply $\frac{1}{2}$ times $1$

$\left(\frac{1}{2}\sqrt{x}\left(x-3\right)^{-\frac{1}{2}}+\frac{1}{2}\sqrt{x-3}x^{-\frac{1}{2}}\right)\frac{1}{\sqrt{x-3}\sqrt{x}}$
10

Multiplying the fraction and term

$\frac{\frac{1}{2}\sqrt{x}\left(x-3\right)^{-\frac{1}{2}}+\frac{1}{2}\sqrt{x-3}x^{-\frac{1}{2}}}{\sqrt{x-3}\sqrt{x}}$
11

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

$\frac{\frac{1}{2}\sqrt{x}\cdot\frac{1}{\sqrt{x-3}}+\frac{1}{2}\sqrt{x-3}\cdot\frac{1}{\sqrt{x}}}{\sqrt{x-3}\sqrt{x}}$
12

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

$\frac{\sqrt{x}\cdot\frac{\frac{1}{2}}{\sqrt{x-3}}+\sqrt{x-3}\cdot\frac{\frac{1}{2}}{\sqrt{x}}}{\sqrt{x-3}\sqrt{x}}$
13

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

$\frac{\sqrt{x}\cdot\frac{\frac{1}{2}}{\sqrt{x-3}}+\frac{1}{2}\sqrt{x-3}x^{-\frac{1}{2}}}{\sqrt{x-3}\sqrt{x}}$
14

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

$\frac{\sqrt{x}\cdot\frac{\frac{1}{2}}{\sqrt{x-3}}+\frac{1}{2}\sqrt{x-3}\cdot\frac{1}{\sqrt{x}}}{\sqrt{x-3}\sqrt{x}}$
15

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

$\frac{\sqrt{x}\cdot\frac{\frac{1}{2}}{\sqrt{x-3}}+\sqrt{x-3}\cdot\frac{\frac{1}{2}}{\sqrt{x}}}{\sqrt{x-3}\sqrt{x}}$

Answer

$\frac{\sqrt{x}\cdot\frac{\frac{1}{2}}{\sqrt{x-3}}+\sqrt{x-3}\cdot\frac{\frac{1}{2}}{\sqrt{x}}}{\sqrt{x-3}\sqrt{x}}$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.52 seconds

Views:

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