Question

Question #b7835

Answer 1

`-8/(2x-1)^2`

Explanation:

`d/dx4/(2x-1)`

`4*d/dx1/(2x-1)`

`4*d/dx(2x-1)^-1`

Remember the chain rule: for `(df(u))/dx = (df)/(du)*(du)/(dx)`

Here, `f=u^-1`, and `u=2x-1`

`(df)/(du)=d/(du) u^-1`

Power rule: `d/dxx^n=nx^(n-1)`

`=-1/u^2`

`(du)/dx=d/dx2x-1`

Remember that `d/dxk`, where `k` is a constant, is equal to `0`, and `d/dx2x=2`.

`=2`

So now we have `4(-1/u^2)2`

But remember, `u=2x-1`, so:

`4(-1/(2x-1)^2)2`

`-8/(2x-1)^2`

Answer 2

`-8/(2x-1)^2`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`4/(2x-1)=4(2x-1)^-1`

`rArrd/dx(4(2x-1)^-1)`

`=-4(2x-1)^-2xxd/dx(2x-1)`

`=-8/(2x-1)^2`