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what is the first and second derivative of (2x-3)(x+4)^-1
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\[f' = \frac{-2x+3}{(x+4)^{2}}+\frac{2}{x+4} = \frac{11}{(x+4)^{2}}\]
\[f'' = \frac{22}{(x+4)^{3}}\]
\[f'' = -\frac{22}{(x+4)^{3}}\] You missed the minus.
It's very unreadable, so I'm just going to say that you're wrong and give you proof: http://www.wolframalpha.com/input/?i=derive+%282x-3%29%28x%2B4%29^-1
how do u get 11?
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Are you familiar with the quotient rule? If we apply that we get for nominator: 2(x+4)-1(2x-3)=11
by multiplying the right side by \[\frac{x+4}{x+4}\] and simplifying after using the product rule, or as thomas said, by the quotient rule.
what about the derivative of \[\sqrt{x}\left( 2x+3 \right)^{2}\]
product rule \[uv' + vu'\]
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