chain rule help, derivative.

**Tweety** · January 3rd 2010, 12:22 PM

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?

**pickslides** · January 3rd 2010, 12:41 PM

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

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 $\frac{dw}{dx}= 2$

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

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

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

**Tweety** · January 3rd 2010, 01:45 PM

oh I completely forgot this step, thank you. I have to differentiate the inside first and than multiply.

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