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Calculating the derivative of a function of the form $\\frac{\\ln(x)}{ax^n}$ using the quotient rule.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Find the derivative of \\[ \\simplify{y=ln(x)/({a}x^{n})}. \\]
", "advice": "If we have a function of the form $y=\\tfrac{u(x)}{v(x)}$, to calculate its derivative we need to use the quotient rule:
\n\\[ \\dfrac{dy}{dx} = \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2}\\,.\\]
\nThis can be split up into steps:
\nFollowing this process, we must first identify $u(x)$ and $v(x)$.
\nAs \\[ \\simplify{y=ln(x)/({a}x^{n})}, \\]
\nlet \\[ u(x) = \\ln(x) \\quad \\text{and} \\quad v(x)=\\simplify{{a}x^{n}}.\\]
\nNext, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:
\n\\[ \\dfrac{du}{dx} = \\dfrac{1}{x}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{a*n}x^{n-1}}.\\]
\nSubstituting these results into the quotient rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:
\n\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2} \\\\ \\\\&\\,=\\dfrac{\\simplify{{a}x^{n}} \\times\\frac{1}{x} - \\ln(x) \\times \\simplify{{n*a}x^{n-1}}}{\\simplify{({a}x^{n})^2}}. \\end{split}\\]
\n\nSimplifying,
\n\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{\\simplify{{a}x^{n-1}-{a*n}ln(x)x^{n-1}}}{\\simplify{{a^2}x^{2*n}}} \\\\ \\\\ &\\,=\\dfrac{\\simplify{{a}x^{n-1}(1-{n}ln(x))}}{\\simplify{{a^2}x^{2n}}}\\\\ \\\\ &\\,=\\dfrac{\\simplify{1-{n}ln(x)}}{\\simplify{{a}x^{n+1}}}. \\end{split} \\]
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