Determine the stationary points of y=(5x^2)/(lnx)
Differentiate y with respect to x using quotient rule:y'=[(1/x)(5x^2)-(10x)(lnx)]/(lnx)^2 =[5x-10xlnx]/(lnx)^2Stationary points occur when y'=0, so when y'=0 we have:5x-10xlnx = 0x(5-10lnx)=0So x=0 or 5-10lnx=0But when x=0, lnx is undefined, so there is no y value at x=0. So x cannot equal 0.Therefore: 5-10lnx=0 x=e^0.5Substitute back into y, we obtain:y=5e/0.5 = 10eSo Sationary Point is: (e^0.5, 10e)
JL
Related Maths A Level answers
All answers ▸Given that log_{x} (7y+1) - log_{x} (2y) =1 x>4, 0<y<1 , express y in terms of x.
What is the integral of x^2 sin(x) between the limits 0 and π/2
Find the total area enclosed between y = x^3 - x, the x axis and the lines x = 1 and x= -1 . (Why do i get 0 as an answer?)
Find the derivative of sin^2(x)