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Thread: chain rule help, derivative.

  1. #1
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    chain rule help, derivative.

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

    $\displaystyle u = x-1 $

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

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

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

    but my books says $\displaystyle \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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    $\displaystyle v = (2x+1)^{-1}$

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

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

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

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

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

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