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We use the PRODUCT RULE when the function that we need to differentiate is actually two functions multiplied together:

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If  $y=u \\times v$  then:

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\\[   \\frac{dy}{dx}=u \\frac{dv}{dx} + v \\frac{du}{dx} \\]

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", "advice": "

We are asked to differentiate:

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\\[ \\large y=x^{\\var{b}} \\ln{(\\var{b}x)} \\]

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Recognising that the function to differentiate is the product of two functions, we identify the two functions that are involved.

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$u$ is the first function, $v$ is second:

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$u=x^{\\var{b}}$                    $v=\\ln{(\\var{b}x)}$

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Now, we need to use the approriate techniques to differentiate each of these, for $u$ we need the Power Rule and $v$ can be done using your Table of Derivatives:

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Applying these methods gives us:

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$\\large \\frac{du}{dx}=\\simplify{ {b} * x^{{{b}}-1} } $          and          $ \\large \\frac{dv}{dx}=\\frac{1}{x}$

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We now use the formula:

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$ \\large  \\frac{dy}{dx}=u \\frac{dv}{dx} + v \\frac{du}{dx} $

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 Make the appropriate substitutions into the formula:

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$ \\large  \\frac{dy}{dx}= x^{\\var{b}} \\times \\frac{1}{x} + \\ln{(\\var{b}x)} \\times \\simplify{{b} *x^{{{b}}-1} } $

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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:

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$ \\large \\frac{dy}{dx}= \\simplify{x^{{b}} * x^(-1)} + \\simplify{{b} *x^{{{b}}-1} } \\ln{(\\var{b}x)}$

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Differentiate  $  y=x^{\\var{b}} \\ln{(\\var{b}x)}$

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First identify the two functions  $u$  and  $v$:

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$u=$[[0]]                    $v=$[[1]]

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Now differentiate each one:

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$  \\large \\frac{du}{dx}=   $[[2]]                    $  \\large\\frac{dv}{dx}=   $[[3]]

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Then using:

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$ \\large  \\frac{dy}{dx}=u \\frac{dv}{dx} + v \\frac{du}{dx} $

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Substitute each component into the formula in the correct place:

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$  \\large \\frac{dy}{dx}=$[[4]]$  \\large \\times$[[5]]$ \\large + $[[6]]$ \\large \\times$[[7]]

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Finally tidy this up to give your final answer:

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$  \\large \\frac{dy}{dx}=   $[[8]]

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