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

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

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Answer

$\frac{\frac{1}{2}\left(2+2x\right)}{2x+x^2}$

Step by step solution

Problem

$\frac{d}{dx}\left(\ln\left(\sqrt{x^2+2x}\right)\right)$
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 $\displaystyle f'(x)=\frac{a'}{a}$

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

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

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

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

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

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

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

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

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

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

$\frac{1}{2}\cdot\frac{1}{\sqrt{2x+x^2}}\left(1\cdot 2+2x\right)\left(2x+x^2\right)^{-\frac{1}{2}}$
7

Multiply $2$ times $1$

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

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

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

Multiplying the fraction and term

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

Simplifying the fraction by $2x+x^2$

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

Subtract the values $-\frac{1}{2}$ and $-\frac{1}{2}$

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

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

$\frac{1}{2}\left(2+2x\right)\left(\frac{1}{2x+x^2}\right)$
13

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

$\left(2+2x\right)\frac{\frac{1}{2}}{2x+x^2}$
14

Multiplying the fraction and term

$\frac{\frac{1}{2}\left(2+2x\right)}{2x+x^2}$

Answer

$\frac{\frac{1}{2}\left(2+2x\right)}{2x+x^2}$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.32 seconds

Views:

98