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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 $\displaystyle f'(x)=\frac{a'}{a}$
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$\frac{1}{\frac{\left(x+2\right)^5}{\sqrt[6]{3x-5}}}\frac{d}{dx}\left(\frac{\left(x+2\right)^5}{\sqrt[6]{3x-5}}\right)$
Learn how to solve product rule of differentiation problems step by step online. Find the derivative using the product rule d/dx(ln(((x+2)^5)/((3x-5)^1/6))). 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}. Divide fractions \frac{1}{\frac{\left(x+2\right)^5}{\sqrt[6]{3x-5}}} with Keep, Change, Flip: a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}. 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}}. Simplify \left(\sqrt[6]{3x-5}\right)^2 using the power of a power property: \left(a^m\right)^n=a^{m\cdot n}. In the expression, m equals \frac{1}{6} and n equals 2.