Derive the function (ln(x^2+4x))/(x^2+4x) with respect to x

\frac{d}{dx}\left(\frac{\ln\left(x^2+4x\right)}{x^2+4x}\right)

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Answer

$\frac{\left(4x+x^2\right)\left(4+2x\right)\frac{1}{4x+x^2}-\left(4+2x\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$

Step by step solution

Problem

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

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\left(4x+x^2\right)\frac{d}{dx}\left(\ln\left(4x+x^2\right)\right)-\frac{d}{dx}\left(4x+x^2\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$
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 $\displaystyle f'(x)=\frac{a'}{a}$

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

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

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

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

$\frac{\left(4x+x^2\right)\frac{1}{4x+x^2}\left(4\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(x^2\right)\right)-\left(4\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(x^2\right)\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$
5

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

$\frac{\left(4x+x^2\right)\frac{1}{4x+x^2}\left(4\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(x^2\right)\right)-\left(1\cdot 4+\frac{d}{dx}\left(x^2\right)\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$
6

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

$\frac{\left(4x+x^2\right)\frac{1}{4x+x^2}\left(1\cdot 4+\frac{d}{dx}\left(x^2\right)\right)-\left(1\cdot 4+\frac{d}{dx}\left(x^2\right)\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$
7

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

$\frac{\left(4x+x^2\right)\left(1\cdot 4+2x\right)\frac{1}{4x+x^2}-\left(1\cdot 4+2x^{\left(2-1\right)}\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$
8

Subtract the values $2$ and $-1$

$\frac{\left(4x+x^2\right)\left(1\cdot 4+2x\right)\frac{1}{4x+x^2}-\left(1\cdot 4+2x^{1}\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$
9

Multiply $4$ times $1$

$\frac{\left(4x+x^2\right)\left(4+2x\right)\frac{1}{4x+x^2}-\left(4+2x^{1}\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$
10

Any expression to the power of $1$ is equal to that same expression

$\frac{\left(4x+x^2\right)\left(4+2x\right)\frac{1}{4x+x^2}-\left(4+2x\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$

Answer

$\frac{\left(4x+x^2\right)\left(4+2x\right)\frac{1}{4x+x^2}-\left(4+2x\right)\ln\left(4x+x^2\right)}{\left(4x+x^2\right)^2}$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.34 seconds

Views:

103