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Math Help - chain rule help, derivative.

  1. #1
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    chain rule help, derivative.

    Find the value of  \frac{dy}{dx} at the point  (2 , \frac{1}{5} ) on the curve with equation  y = (x-1)(2x+1)^{-1}

     u = x-1

     v = (2x+1)^{-1}

    I know you have to use the chain rule on v so this is what I get,

     \frac{dv}{dx} = -1 \times (2x+1)^{-2}

    but my books says  \frac{dv}{dx} = -2(2x+1)^{-2}

    How have they got -2? As I have been taught that you first bring down the power to the front of the expression and than minus 1 from it.

    Is this not correct?
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  2. #2
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     v = (2x+1)^{-1}

    Make w = 2x+1 so  v = w^{-1}

    Now  \frac{dv}{dx}= \frac{dv}{dw}\times \frac{dw}{dx}

     \frac{dv}{dw} = -w^{-2} and \frac{dw}{dx}= 2

     \frac{dv}{dx}= \frac{dv}{dw}\times \frac{dw}{dx}

     \frac{dv}{dx}=  -w^{-2}\times 2

     \frac{dv}{dx}=  -2(2x+1)^{-2}
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  3. #3
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    oh I completely forgot this step, thank you. I have to differentiate the inside first and than multiply.
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