# Derive the function ln(x(a+x)^0.5) with respect to x

## \frac{d}{dx}\left(\ln\left(x\sqrt{a+x}\right)\right)

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$\frac{x\frac{\frac{1}{2}}{\sqrt{x+a}}+\sqrt{x+a}}{x\sqrt{x+a}}$

## Step by step solution

Problem

$\frac{d}{dx}\left(\ln\left(x\sqrt{a+x}\right)\right)$
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 $\displaystyle f'(x)=\frac{a'}{a}$

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

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

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

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

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

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

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

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

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

The derivative of the constant function is equal to zero

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

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

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

Add the values $0$ and $1$

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

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

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

Any expression multiplied by $1$ is equal to itself

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

Multiplying the fraction and term

$\frac{\frac{1}{2}x\left(x+a\right)^{-\frac{1}{2}}+\sqrt{x+a}}{x\sqrt{x+a}}$
12

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

$\frac{\frac{1}{2}x\frac{1}{\sqrt{x+a}}+\sqrt{x+a}}{x\sqrt{x+a}}$
13

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

$\frac{x\frac{\frac{1}{2}}{\sqrt{x+a}}+\sqrt{x+a}}{x\sqrt{x+a}}$

$\frac{x\frac{\frac{1}{2}}{\sqrt{x+a}}+\sqrt{x+a}}{x\sqrt{x+a}}$

### Main topic:

Differential calculus

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