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Step-by-step Solution

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

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Answer

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

Step-by-step explanation

Problem to solve:

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

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

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

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

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Answer

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

Main topic:

Differential calculus

Used formulas:

6. See formulas

Time to solve it:

~ 0.84 seconds