Question

How do you differentiate `f(x)=(2x)/(sqrt(5x^2 -2x + 1))` using the quotient rule?

Answer 1

`-(2(x-1))/(5x^2-2x+1)^(3/2)`

Explanation:

According to the quotient rule,
`f'(x)=(d/(dx)[2x]*(5x^2-2x+1)^(1/2)-2x*d/(dx)[(5x^2-2x+1)^(1/2)])/(5x^2-2x+1)`

Let's first calculate the two derivatives inside.
`d/(dx)[2x]=2`

The second requires the chain rule.
`d/(dx)[(5x^2-2x+1)^(1/2)]`
`=1/2(5x^2-2x+1)^(-1/2)*d/(dx)[5x^2-2x+1]`
`=(10x-2)/(2(5x^2-2x+1)^(1/2))=(5x-1)/(5x^2-2x+1)^(1/2)`

Plug those back in.
`f'(x)=(2(5x^2-2x+1)^(1/2)-(2x(5x-1))/(5x^2-2x+1)^(1/2))/(5x^2-2x+1)`
`=(2(5x^2-2x+1)-2x(5x-1))/(5x^2-2x+1)^(3/2)`
`=(cancel(10x^2)-4x+2cancel(-10x^2)+2x)/(5x^2-2x+1)^(3/2)`
`=color(blue)(-(2(x-1))/(5x^2-2x+1)^(3/2)`