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Calculating the derivative of a function of the form $ax^n(bx^m+c)^p$ using the product rule.

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Find the derivative of \\[ \\simplify{y={a}x^{n}({b}x^{m}+{c})^{p}}. \\]

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If we have a function of the form $y=u(x)v(x)$, to calculate its derivative we need to use the product rule:

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\\[ \\dfrac{dy}{dx} = u(x) \\times \\dfrac{dv}{dx} + v(x) \\times\\dfrac{du}{dx}.\\]

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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 product rule to obtain an expression for $\\tfrac{dy}{dx}$;
  6. \n
  7. Simplify $\\tfrac{dy}{dx}$ where possible.
  8. \n
\n

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

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As \\[ \\simplify{y={a}x^{n}({b}x^{m}+{c})^{p}}, \\]

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let \\[ u(x) = \\simplify{{a}x^{n}} \\quad \\text{and} \\quad v(x)=\\simplify{({b}x^{m}+{c})^{p}}.\\]

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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*n}x^{n-1}}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{b*p*m}x^{m-1}({b}x^{m}+{c})^{p-1}}.\\]

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

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\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{du}{dx}\\times v(x) + u(x) \\times\\dfrac{dv}{dx} \\\\ \\\\&\\,=\\simplify{{a*n}x^{n-1}} \\times\\simplify{({b}x^{m}+{c})^{p}} +\\simplify{{a}x^{n}} \\times \\simplify{{b*p*m}x^{m-1}({b}x^{m}+{c})^{p-1}}.  \\end{split}\\]

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Simplifying,

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\\\\[ \\\\begin{split} \\\\dfrac{dy}{dx} &\\\\,= \\\\simplify{{a*n}x^{n-1}({b}x^{m}+{c})^{p}+{a}x^{n}{b*p*m}x^{m-1}({b}x^{m}+{c})^{p-1}} \\\\\\\\ &\\\\,= \\\\simplify[all,!expandBrackets,!cancelTerms]{{simp2}x^{n-1}({b}x^{m}+{c})^{p-1}}\\\\big(\\\\simplify[all,!expandBrackets,!cancelTerms]{{a*n/simp2}({b}x^{m}+{c})+{a*b*p*m/simp2}x^{m}}\\\\big)\\\\\\\\ &\\\\,=\\\\simplify[all]{{a}x^{n-1}({b}x^{m}+{c})^{p-1}({n*b+b*p*m}x^{m}+{n*c})}\\\\\\\\ &\\\\,=\\\\simplify{{a*simp}x^{n-1}({b}x^{m}+{c})^{p-1}({(n*b+b*p*m)/simp}x^{m}+{n*c/simp})}.\\\\end{split} \\\\]

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Simplifying,

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\\\\[ \\\\begin{split} \\\\dfrac{dy}{dx} &\\\\,= \\\\simplify{{a*n}x^{n-1}({b}x^{m}+{c})^{p}+{a}x^{n}{b*p*m}x^{m-1}({b}x^{m}+{c})^{p-1}} \\\\\\\\ &\\\\,= \\\\simplify{{a}x^{n-1}({b}x^{m}+{c})^{p-1}}\\\\big(\\\\simplify{{n}({b}x^{m}+{c})+{b*p*m}x^{m}}\\\\big)\\\\\\\\ &\\\\,=\\\\simplify[all]{{a}x^{n-1}({b}x^{m}+{c})^{p-1}({n*b+b*p*m}x^{m}+{n*c})}.\\\\end{split} \\\\]

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Simplifying,

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\\\\[ \\\\begin{split} \\\\dfrac{dy}{dx} &\\\\,= \\\\simplify{{a*n}x^{n-1}({b}x^{m}+{c})^{p}+{a}x^{n}{b*p*m}x^{m-1}({b}x^{m}+{c})^{p-1}} \\\\\\\\ &\\\\,= \\\\simplify[all,!expandBrackets,!cancelTerms]{{simp2}x^{n-1}({b}x^{m}+{c})^{p-1}}\\\\big(\\\\simplify[all,!expandBrackets,!cancelTerms]{{a*n/simp2}({b}x^{m}+{c})+{a*b*p*m/simp2}x^{m}}\\\\big)\\\\\\\\&\\\\,=\\\\simplify[all]{{simp2}x^{n-1}({b}x^{m}+{c})^{p-1}({(a*n*b+a*b*p*m)/simp2}x^{m}+{a*n*c/simp2})}\\\\\\\\ &\\\\,=\\\\simplify{{a*simp}x^{n-1}({b}x^{m}+{c})^{p-1}({(n*b+b*p*m)/simp}x^{m}+{n*c/simp})}.\\\\end{split} \\\\]

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Simplifying,

\\n

\\\\[ \\\\begin{split} \\\\dfrac{dy}{dx} &\\\\,= \\\\simplify{{a*n}x^{n-1}({b}x^{m}+{c})^{p}+{a}x^{n}{b*p*m}x^{m-1}({b}x^{m}+{c})^{p-1}} \\\\\\\\ &\\\\,= \\\\simplify[all,!expandBrackets,!cancelTerms]{{simp2}x^{n-1}({b}x^{m}+{c})^{p-1}}\\\\big(\\\\simplify[all,!expandBrackets,!cancelTerms]{{a*n/simp2}({b}x^{m}+{c})+{a*b*p*m/simp2}x^{m}}\\\\big)\\\\\\\\&\\\\,=\\\\simplify[all]{{simp2}x^{n-1}({b}x^{m}+{c})^{p-1}({(a*n*b+a*b*p*m)/simp2}x^{m}+{a*n*c/simp2})} .\\\\end{split} \\\\]

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$\\dfrac{dy}{dx}=$[[0]] 

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