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We are asked to differentiate:
\n\\[ y=\\var{bCF}x^{\\var{bP1}} \\cos{(\\var{bCF2}x)}\\]
\n\nRecognising that the function to differentiate is the product of two functions, we identify the two functions that are involved.
\n\n$u$ is the first function, $v$ is second:
\n\n$\\large u=\\var{bCF}x^{\\var{bP1}} $ $\\large v=\\cos{(\\var{bCF2}x)} $
\n\n
Now, we need to use the approriate techniques to differentiate each of these, for $u$ we can use the Power Rule and for $v$ your Table of Derivatives:
\n\nThis gives us:
\n$\\large \\frac{du}{dx}=\\simplify{ {bP1}*{bCF}*x^({bP1}-1) }$ and $ \\large \\frac{dv}{dx}=- \\var{bCF2} \\sin{(\\var{bCF2}x)}$
\n\nWe now use the formula:
\n$ \\large \\frac{dy}{dx}=u \\frac{dv}{dx} + v \\frac{du}{dx} $
\nMake the appropriate substitutions into the formula:
\n\n$ \\large \\frac{dy}{dx}= \\var{bCF}x^{\\var{bP1}} \\times - \\var{bCF2}\\sin{(\\var{bCF2}x)} + \\cos{(\\var{bCF2}x)} \\times \\simplify{ {bP1}*{bCF}*x^({bP1}-1) } $
\n\n\n
Finally, we need to use our basic algebra to simplify this as much as possible. Multiply out any brackets where it would simplify and collect like terms:
\n\n$ \\large \\frac{dy}{dx}= \\simplify{ - {bCF2}* {bCF} x^{{bP1}}} \\sin{(\\var{bCF2}x)} + \\simplify{ {bP1}*{bCF}*x^({bP1}-1) } \\cos{(\\var{bCF2}x)} $
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First identify the two functions $u$ and $v$:
\n$u=$[[0]] $v=$[[1]]
\n\n
Now differentiate each one:
\n$ \\large \\frac{du}{dx}= $[[2]] $ \\large\\frac{dv}{dx}= $[[3]]
\n\n
Substitute each component into the formula in the correct place:
\n$ \\large \\frac{dy}{dx}=$[[4]]$ \\large \\times$[[5]]$ \\large + $[[6]]$ \\large \\times$[[7]]
\n\n
Finally tidy this up to give your final answer:
\n$ \\large \\frac{dy}{dx}= $[[8]]
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