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Differentiate the function $f(x)=(a + b x)^m  e ^ {n x}$ using the product rule. Find $g(x)$ such that $f^{\\prime}(x)= (a + b x)^{m-1}  e ^ {n x}g(x)$.

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Differentiate the following function $f(x)$ using the product rule.

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The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]

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For this example:

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\\[\\simplify[dPoly]{u = ({a} + {b} * x) ^ {m}}\\Rightarrow \\simplify[dPoly]{Diff(u,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1}}\\]

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\\[\\simplify{v = e ^ ({n} * x)} \\Rightarrow \\simplify{Diff(v,x,1) = {n} * e ^ ({n} * x)}\\]

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Hence on substituting into the product rule above we get:

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\\[\\simplify[dPoly]{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x) = ({a} + {b} * x) ^ {m -1} * ({m * b + n * a} + {n * b} * x) * e ^ ({n} * x)}\\]

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The last step was to take out the common term $\\simplify[dPoly]{({a} + {b} * x) ^ {m -1} * e ^ ({n} * x)}$.

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Hence \\[\\simplify[dPoly]{g(x) = {m * b + n * a} + {n * b} * x}\\].

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$\\simplify[dPoly]{f(x) = ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$

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You are given that \\[\\simplify[dPoly]{Diff(f,x,1) = ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) * g(x)}\\]

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for a polynomial $g(x)$. You have to find $g(x)$.

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$g(x)=\\;$[[0]]

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Click on Show steps to get more information, you will not lose any marks by doing so.

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The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]

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