# Thread: chain rule help, derivative.

1. ## 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?

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

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

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

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

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

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