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Calculating the derivative of a function of the form $\\frac{(x^2-a)^n}{x^2+a}$ using the quotient rule.

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Find the derivative of \\[ \\simplify{y=(x^2-{a})^{n}/(x^2+{a})}. \\]

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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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    \n
  1. Identify the functions $u(x)$ and $v(x)$;
  2. \n
  3. Calculate their derivatives $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$;
  4. \n
  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
\n

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Following this process, we must first identify $u(x)$ and $v(x)$.

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As \\[ \\simplify{y=(x^2-{a})^{n}/(x^2+{a})}, \\]

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let \\[ u(x) = \\simplify{(x^2-{a})^{n}} \\quad \\text{and} \\quad v(x)=\\simplify{x^2+{a}}.\\]

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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{{2*n}x(x^2-{a})^{n-1}}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=2x.\\]

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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{x^2+{a}}) \\times \\simplify{{2*n}x(x^2-{a})^{n-1}} - \\simplify{(x^2-{a})^{n}} \\times \\simplify{2x}}{\\simplify{(x^2+{a})^2}}.  \\end{split}\\]

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Simplifying,

\\n

\\\\[ \\\\begin{split} \\\\dfrac{dy}{dx} &\\\\,= \\\\dfrac{\\\\simplify{{2*n}x(x^2+{a})(x^2-{a})^{n-1}-2x(x^2-{a})^{n}}}{\\\\simplify{(x^2+{a})^2}}  \\\\\\\\ \\\\\\\\ &\\\\,= \\\\dfrac{\\\\simplify{2x(x^2-{a})^{n-1}({n}(x^2+{a})-(x^2-{a}))}}{\\\\simplify{(x^2+{a})^2}}\\\\\\\\ \\\\\\\\ &\\\\,= \\\\dfrac{\\\\simplify{2x(x^2-{a})^{n-1}({n-1}x^2+{a*(n+1)})}}{\\\\simplify{(x^2+{a})^2}}. \\\\end{split} \\\\]

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Simplifying,

\\n

\\\\[ \\\\begin{split} \\\\dfrac{dy}{dx} &\\\\,= \\\\dfrac{\\\\simplify{{2*n}x(x^2+{a})(x^2-{a})^{n-1}-2x(x^2-{a})^{n}}}{\\\\simplify{(x^2+{a})^2}}  \\\\\\\\ \\\\\\\\ &\\\\,= \\\\dfrac{\\\\simplify{2x(x^2-{a})^{n-1}({n}(x^2+{a})-(x^2-{a}))}}{\\\\simplify{(x^2+{a})^2}}\\\\\\\\ \\\\\\\\ &\\\\,= \\\\dfrac{\\\\simplify{2x(x^2-{a})^{n-1}({n-1}x^2+{a*(n+1)})}}{\\\\simplify{(x^2+{a})^2}}\\\\\\\\ \\\\\\\\ &\\\\,= \\\\dfrac{\\\\simplify{{2*simp}x(x^2-{a})^{n-1}({(n-1)/simp}x^2+{(a*(n+1))/simp})}}{\\\\simplify{(x^2+{a})^2}}. \\\\end{split} \\\\]

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$\\dfrac{dy}{dx}=$[[0]] 

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