dunno....some guy doing the course i want to do at oxford was asked it during the interview....but he was doing further maths.....i havent ecountered this type of question on p3....and i'm not doing further maths...

Reply 4

M Safe

let y=x^2

(1/y)y'=2xlnx + x (d/dx on both sides)

=> y'= x^(x^2)[2xlnx+x]

sorry y=x^(x^2)

Reply 5

matouwah

M Safe

let y=x^2

(1/y)y'=2xlnx + x (d/dx on both sides)

=> y'= x^(x^2)[2xlnx+x]

uhmmm.....kinda lost....could u explain it in more detail...i am a bit of a retard.....thanx...

Reply 6

M Safe

matouwah

uhmmm.....kinda lost....could u explain it in more detail...i am a bit of a retard.....thanx...

y=x^(x^2)

ln y = (x^2)ln(x)

differentiate wrt x on both sides (dy/dx = d/dy * dy/dx)

so 1/y * (dy/dx) = x + 2xln(x) (use product rule to obtain RHS)

do dy/dx = y(x+2xln(x))
= ...

Reply 7

matouwah

M Safe

y=x^(x^2)

ln y = (x^2)ln(x)

differentiate wrt x on both sides (dy/dx = d/dy * dy/dx)

so 1/y * (dy/dx) = x + 2xln(x) (use product rule to obtain RHS)

do dy/dx = y(x+2xln(x))
= ...

ahhhhh i see it now.....thanx...

take logs.

i once differentiated x^x^x "for fun". woohoo the hilarity! (i was bored in a free period). i then went on to do x^x^x^x, and their integrals. god i am sad.