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Now it is time to get out your paper and pencil and try similar questions without the help. Carry out the same steps, lay it out the same way but you only need to input the final answer.
", "advice": "We are asked to differentiate:
\n\\[ y=\\frac{e^{\\var{b}x}}{x^{\\var{a}}} \\]
\n\nAt first this function looks like two functions divided rather than multiplied, making it a candidate for the Quotient Rule instead of the Product Rule.
\nHowever, recognising that the function $\\frac{1}{x^{\\var{a}}}$ (the denominator) can be rewritten as $x^{-\\var{a}}$ gives us:
\n\\[ y=e^{\\var{b}x} x^{-\\var{a}}\\]
\nWe are now on much more familiar ground for the Product Rule.
\n\n$u$ is the first function, $v$ is second:
\n\n$\\large u=e^{\\var{b}x} $ $ \\large v=x^{-\\var{a}}$
\n\n
Now, we need to use the approriate techniques to differentiate each of these, for these functions we need your Table of Derivatives:
\n\nThis gives us:
\n$\\large \\frac{du}{dx}=\\var{b}e^{\\var{b}x} $ and $ \\large \\frac{dv}{dx}= \\simplify{ -{a}x^{-{a}-1} }$
\n\nWe now use the formula:
\n$ \\large \\frac{dy}{dx}=u \\frac{dv}{dx} + v \\frac{du}{dx} $
\n\nMake the appropriate substitutions into the formula:
\n\n$ \\large \\frac{dy}{dx}= e^{\\var{b}x} \\times \\simplify{-{a}x^{-{a}-1} } + x^{-\\var{a}} \\times \\var{b}e^{\\var{b}x} $
\n\n\n
Finally, we need to use our basic algebra to simplify this as much as possible. Multiply out brackets where it would simplify and collect like terms:
\n\n$ \\large \\frac{dy}{dx}= -\\frac{\\var{a} e^{\\var{b}x} }{\\simplify{x^{{a}+1}}}+\\frac{\\var{b}e^{\\var{b}x}}{x^{\\var{a}}}$
\n\nNotice that, whenever possible, your final answer should not contain negative indices (powers).
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aP
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