# Step-by-step Solution

## Find the derivative using the quotient rule $\frac{d}{dx}\left(\frac{\ln\left(\left(x+2\right)^2\right)\ln\left(\sqrt{x-1}\right)}{\ln\left(\left(x-5\right)^3\right)}\right)$

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### Videos

$\frac{\ln\left(\frac{1}{\left(x-1\right)^{\ln\left(\left(x-5\right)^{24}\right)}\left(x+2\right)^{\ln\left(\left(x-1\right)^{9}\right)}}\right)}{9\ln\left(x-5\right)^2\left(-3x+x^2-10\right)}$

## Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{\left(\ln\left(\left(x+2\right)^2\right)\ln\left(\sqrt{x-1}\right)\right)}{\ln\left(\left(x-5\right)^3\right)}\right)$
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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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\frac{d}{dx}\left(\ln\left(\left(x+2\right)^2\right)\ln\left(\sqrt{x-1}\right)\right)\ln\left(\left(x-5\right)^3\right)-\ln\left(\left(x+2\right)^2\right)\ln\left(\sqrt{x-1}\right)\frac{d}{dx}\left(\ln\left(\left(x-5\right)^3\right)\right)}{\ln\left(\left(x-5\right)^3\right)^2}$
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Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=\ln\left(\left(x+2\right)^2\right)$ and $g=\ln\left(\sqrt{x-1}\right)$

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

$\frac{\ln\left(\frac{1}{\left(x-1\right)^{\ln\left(\left(x-5\right)^{24}\right)}\left(x+2\right)^{\ln\left(\left(x-1\right)^{9}\right)}}\right)}{9\ln\left(x-5\right)^2\left(-3x+x^2-10\right)}$
$\frac{d}{dx}\left(\frac{\left(\ln\left(\left(x+2\right)^2\right)\ln\left(\sqrt{x-1}\right)\right)}{\ln\left(\left(x-5\right)^3\right)}\right)$