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Other method. Find $p,\\;q$ such that $\\displaystyle \\frac{ax+b}{cx+d}= p+ \\frac{q}{cx+d}$. Find the derivative of $\\displaystyle \\frac{ax+b}{cx+d}$.
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\\[\\simplify[std]{f(x) = a + b/({c}x+{d})}\\]
Enter a and b as integers or fractions, but not as decimals.
$a=\\;$[[0]]
\n\t\t\t$b=\\;$[[1]]
\n\t\t\tYou can click on Show steps to get some help, but you will lose 1 mark if you do so.
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\\[\\simplify[std]{g(x) = 1/({c}x+{d})}\\]
$\\displaystyle \\frac{dg}{dx}=\\;$[[0]]
\n\t\t\tHence using the first part of the question differentiate \\[\\simplify[std]{f(x) = ({a} * x+{b})/({c}x+{d})}\\]
\n\t\t\t$\\displaystyle \\frac{df}{dx}=\\;$[[1]]
\n\t\t\tInput numbers as fractions or integers and not as decimals.
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\n\tWe have $\\displaystyle \\simplify[std]{{a}x+{b}={a}/{c}*({c}x+{d})+{b}-{a}*{d}/{c}={a}/{c}*({c}x+{d})+{-det}/{c}}$
Hence \\[\\begin{eqnarray*} \\simplify[std]{({a} * x+{b})/({c}x+{d})}&=&\\simplify[std]{({a}/{c}*({c}x+{d})+{-det}/{c})/({c}x+{d})}\\\\ &=&\\simplify[std]{{a}/{c}+({-det}/{c})/({c}x+{d})} \\end{eqnarray*}\\]
Where we have divided out by $\\simplify[std]{{c}x+{d}}$ at the last step.
b)
\n\tWe have \\[\\frac{dg}{dx} = \\simplify[std]{{-c}/({c}x+{d})^2}\\]
using standard rules of differentiation.
Since from a), \\[f(x) = \\simplify[std]{{a}/{c}+({-det}/{c})/({c}x+{d})}\\]
we see that
\\[\\begin{eqnarray*}\\frac{df}{dx} &=&\\simplify[std,!unitPower,!unitDenominator,!zeroFactor,!zeroTerm,!zeroPower]{(-{c})*(({-det}/{c})/({c}x+{d})^2)}\\\\ &=&\\simplify[std]{{det}/({c}x+{d})^2} \\end{eqnarray*}\\]