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Calculating the derivative of a function of the form $\\frac{a\\sin(x)}{bx+c \\cos(x)}$ using the quotient rule.

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Find the derivative of \\[ \\simplify{y={a}sin(x)/({b}x+{c}cos(x))}. \\]

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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:

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\\[ \\dfrac{dy}{dx} = \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2}\\,.\\]

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This can be split up into steps:

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  1. Identify the functions $u(x)$ and $v(x)$;
  2. \n
  3. Calculate their derivatives $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$;
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  5. Substitute these into the formula for the quotient rule to obtain an expression for $\\tfrac{dy}{dx}$;
  6. \n
  7. Simplify $\\tfrac{dy}{dx}$ where possible.
  8. \n
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Following this process, we must first identify $u(x)$ and $v(x)$.

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As\\[ \\simplify{y={a}sin(x)/({b}x+{c}cos(x))}, \\]

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let \\[ u(x) = \\simplify{{a}sin(x)} \\quad \\text{and} \\quad v(x)=\\simplify{{b}x+{c}cos(x)}.\\]

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Next, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:

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\\[ \\dfrac{du}{dx} = \\simplify{{a}cos(x)}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{b}-{c}sin(x)}.\\]

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Substituting these results into the quotient rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:

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\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2} \\\\ \\\\&\\,=\\dfrac{(\\simplify{{b}x+{c}cos(x)}) \\times\\simplify{{a}cos(x)} - \\simplify{{a}sin(x)} \\times (\\simplify{{b}-{c}sin(x)})}{\\simplify{({b}x+{c}cos(x))^2}}.  \\end{split}\\]

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Simplifying,

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\\[ \\begin{split} \\dfrac{dy}{dx} &\\,=\\dfrac{(\\simplify{{b}x+{c}cos(x)})\\simplify{{a}cos(x)} - \\simplify{{a}sin(x)}(\\simplify{{b}-{c}sin(x)})}{\\simplify{({b}x+{c}cos(x))^2}} \\\\ \\\\&\\,=\\dfrac{\\simplify[all,!cancelTerms]{{a*b}x cos(x) +{a*c}cos(x)^2-{a*b}sin(x)+{a*c}sin(x)^2}}{\\simplify{({b}x+{c}cos(x))^2}} \\\\ \\\\ &\\,=\\dfrac{\\simplify[all,!cancelTerms]{{a*b}(x cos(x)-sin(x)) +{a*c}(cos(x)^2+sin(x)^2)}}{\\simplify{({b}x+{c}cos(x))^2}} \\\\ \\\\ &\\,=\\dfrac{\\simplify[all,!cancelTerms]{{a*b}(x cos(x)-sin(x)) +{a*c}}}{\\simplify{({b}x+{c}cos(x))^2}} \\end{split} \\]

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$\\dfrac{dy}{dx}=$[[0]] 

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