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

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

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You are told that $\\simplify{Diff(f,x,1) = ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) * g(x)}$, for a polynomial $g(x)$.

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You have to find $g(x)$.

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

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$f(x)$ is the product of the two functions $\\simplify{({a} + {b}*x)^{m}}$ and $\\simplify{e ^ ({n} * x)}$, so we need to use the product rule.

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Differentiating the first part, keeping the second half the same, gives the term: $\\simplify{{m} *{ b} * ({a} + {b} * x) ^ {m -1}} \\times \\simplify{e ^ ({n} * x)}$.  

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Note that that we needed the chain rule to do this differentiation.

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Differentiating the second part, keeping the first half the same, gives the term: $\\simplify{{n} * e ^ ({n} * x)} \\times \\simplify{({a} + {b}x)^{m}}$.

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Again, we needed the chain rule to do this differentiation.

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Hence, $\\simplify{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$.

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           $= \\simplify{({a} + {b} * x) ^ {m -1} * ({m * b + n * a} + {n * b} * x) * e ^ ({n} * x)}$, (by doing some factorising)

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Hence, $\\simplify{g(x) = {m * b + n * a} + {n * b} * x}$.

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Differentiate the following function $f(x)$.

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