A function is defined as f(x) = x / sqrt(2x-2). Use the quotient rule to show that f'(x) = (x-2)/(2x-2)^(3/2)

u = x v = (2x-2)^(0.5)u' = 1 v' = (2x-2)^(-0.5)f'(x) = (vu' - uv') / v^2Therefore, f'(x) = (((2x-2)^(0.5) * 1) - (x * (2x-2)^(-0.5))) / ((2x-2)^(0.5))^2f'(x) = (2x - 2 - x) / (2x-2)^(3/2) = (x-2) / (2x-2)^(3/2)Would be easier to follow with the whiteboard function

Answered by Isaac F. Maths tutor

6832 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the equation of the normal to the curve at the point (1, -1 ): 10yx^2 + 6x - 2y + 3 = x^3


Find values of y such that: log2(11y–3)–log2(3) –2log2(y) = 1


How do I differentiate: (3x + 7)^2?


Find the tangent to the curve y = x^2 + 3x + 2 that passes through the point (-1,0), sketch the curve and the tangent.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy