Find the derivative of ln((x^2+1)/(x+3))^0.5

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

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Answer

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

Step by step solution

Problem

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

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

$\frac{1}{2}\cdot\frac{d}{dx}\left(\ln\left(\frac{1+x^2}{3+x}\right)\right)\ln\left(\frac{1+x^2}{3+x}\right)^{-\frac{1}{2}}$
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 $\displaystyle f'(x)=\frac{a'}{a}$

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

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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

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

The derivative of the constant function is equal to zero

$\frac{1}{2}\left(\frac{1}{\frac{1+x^2}{3+x}}\right)\ln\left(\frac{1+x^2}{3+x}\right)^{-\frac{1}{2}}\left(\frac{\left(3+x\right)\left(0+\frac{d}{dx}\left(x^2\right)\right)-\left(1+x^2\right)\left(0+\frac{d}{dx}\left(x\right)\right)}{\left(3+x\right)^2}\right)$
6

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

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

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

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

Add the values $1$ and $0$

$\frac{1}{2}\left(\frac{1\left(-1\right)\left(1+x^2\right)+\left(3+x\right)\left(0+2x\right)}{\left(3+x\right)^2}\right)\left(\frac{1}{\frac{1+x^2}{3+x}}\right)\ln\left(\frac{1+x^2}{3+x}\right)^{-\frac{1}{2}}$
9

Multiply $-1$ times $1$

$\frac{1}{2}\left(\frac{\left(3+x\right)\left(0+2x\right)-\left(1+x^2\right)}{\left(3+x\right)^2}\right)\left(\frac{1}{\frac{1+x^2}{3+x}}\right)\ln\left(\frac{1+x^2}{3+x}\right)^{-\frac{1}{2}}$
10

$x+0=x$, where $x$ is any expression

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

Simplifying the fraction

$\frac{1}{2}\cdot 1\left(\frac{2x\left(3+x\right)-\left(1+x^2\right)}{\left(3+x\right)^2}\right)\left(\frac{3+x}{1+x^2}\right)\ln\left(\frac{1+x^2}{3+x}\right)^{-\frac{1}{2}}$
12

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

$\frac{1}{2}\cdot\frac{2x\left(3+x\right)-\left(1+x^2\right)}{\left(3+x\right)^2}\cdot\frac{3+x}{1+x^2}\ln\left(\frac{1+x^2}{3+x}\right)^{-\frac{1}{2}}$
13

Split the fraction $\frac{x^2+1}{3+x}$ in two terms with same denominator

$\frac{1}{2}\cdot\frac{2x\left(3+x\right)-\left(1+x^2\right)}{\left(3+x\right)^2}\cdot\frac{3+x}{1+x^2}\ln\left(\frac{1}{3+x}+\frac{x^2}{3+x}\right)^{-\frac{1}{2}}$
14

Add fraction's numerators with common denominators: $\frac{x^2}{3+x}$ and $\frac{1}{3+x}$

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

Answer

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

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.42 seconds

Views:

101