How would I differentiate cos(2x)/x^1/2

So for this question you can use either the product rule or the quotient rule and I'll run through them both.First the quotient rule:The quotient rule says that if you have h(x)=f(x)/g(x)Then h'(x) = (f'(x)g(x)-f(x)g'(x))/(g(x))^2So using f(x)=cos(2x) and g(x)=x^1/2then f'(x)=-2sin(2x) and g'(x)=1/2x^-1/2Plugging this into our formula gives ush(x) = (-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xAlways remember to simplify afterwards which gives us(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xSecond the product rule:What the product rule says is that ifh(x) = f(x)g(x)then h'(x) = f(x)g'(x) + f'(x)g(x)So if we say that h(x) = cos(2x)/x^1/2Then we can say that f(x) = cos(2x) and g(x) = x^-1/2Using the product rule we have:f(x) = cos(2x)      f'(x) = -2sin(2x)g(x) = x^-1/2     g'(x) = 1/2x^-3/2So lastly we know that h(x) = f(x)g'(x) + f'(x)g(x)So using what we've found out we can say that h(x) = (cos(2x))/(2x^3/2)-(2sin(2x))/x^1/2Once again simplifying gives us(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

Answered by George S. Maths tutor

8751 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given f(x) = 3 - 5x + x^3, how can I show that f(x) = 0 has a root (x=a) in the interval 1<a<2?


How can the y=sin(x) graph be manipulated?


The curve has equation y = x^3 - x^2 - 5x + 7 and the straight line has equation y = x + 7. One point of intersection, B, has coordinates (0, 7). Find the other two points of intersection, A and C.


What method should I use to differentiate equations with an x as the power of a number. E.g. 2^x


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences