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Apply the quotient rule for differentiation, 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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
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$\frac{\frac{d}{dx}\left(\ln\left(\left(x+2\right)^2\right)\ln\left(\sqrt{x-1}\right)\right)\ln\left(\left(x-5\right)^3\right)-\frac{d}{dx}\left(\ln\left(\left(x-5\right)^3\right)\right)\ln\left(\left(x+2\right)^2\right)\ln\left(\sqrt{x-1}\right)}{\ln\left(\left(x-5\right)^3\right)^2}$
Learn how to solve problems step by step online. Find the derivative using the quotient rule d/dx((ln((x+2)^2)ln((x-1)^1/2))/ln((x-5)^3)). Apply the quotient rule for differentiation, 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}}. Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=. 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}. Multiplying the fraction by -1.