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)

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Answered by Jimmy L. Maths tutor

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