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Calculating the derivative of a function of the form $(ax+b)^n(cx+d)^m$ using the product rule.

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

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

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

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

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

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Simplifying,

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

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$\\dfrac{dy}{dx}=$[[0]] 

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