How to find Higher Derivatives?

Start by finding the first derivative. You can use the quotient rule for this. Call the numerator f(x) = x and call the denominator g(x) = 2x-1. Then the derivatives f'(x) = 1 and g'(x) = 2. By the quotient rule,

y' = [g(x)f'(x)-f(x)g'(x)]/(g(x))^2

y' = [(2x-1)f'(x) - (x)g'(x)]/(2x-1)^2

y' = [(2x-1)(1) - (x)(2)]/(2x-1)^2

y' = [2x-1 - 2x]/(2x-1)^2

y' = -1/(2x-1)^2

Now find the second derivative. Use the quotient rule again (you could also use the chain rule if you know it.) This time let f(x) = -1 and g(x) = (2x-1)^2=(2x-1)(2x-1)=4x^2-4x+1. Then the derivatives are f'(x) = 0 and g'(x) = 8x-4. By the quotient rule,

y'' = [g(x)f'(x)-f(x)g'(x)]/(g(x))^2

y'' = [(2x-1)^2f'(x)-(-1)g'(x)]/((2x-1)^2)^2

y'' = [(2x-1)^2f'(x)+g'(x)]/(2x-1)^4

y'' = [(2x-1)^2(0)+(8x-4)]/(2x-1)^4

y'' = (8x-4)/(2x-1)^4

y'' = 4(2x-1)/(2x-1)^4

y'' = 4/(2x-1)^3

Now find the third derivative. Use the quotient rule again (or the chain rule.) This time let f(x) = 4 and g(x) = (2x-1)^3=(2x-1)(4x^2-4x+1)=8x^3-8x^2+2x-4x^2+4x-1=8x^3-12x^2+6x-1. Then the derivatives are f'(x) = 0 and g'(x) = 24x^2-24x+6. By the quotient rule,

y''' = [g(x)f'(x)-f(x)g'(x)]/(g(x))^2

y''' = [(2x-1)^3f'(x)-(4)g'(x)]/((2x-1)^3)^2

y''' = [(2x-1)^3f'(x)-4g'(x)]/(2x-1)^6

y''' = [(2x-1)^3(0)-4(24x^2-24x+6)]/(2x-1)^6

y''' = -4(24x^2-24x+6)/(2x-1)^6

y''' = -4*6(4x^2-4x+1)/(2x-1)^6

y''' = -24(4x^2-4x+1)/(2x-1)^6

y''' = -24(2x-1)^2/(2x-1)^6

y''' = -24/(2x-1)^4

Hope this helps you!