# Maths question

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#1
a) Defferenciate x^(x^2) with respect to x

is this **** done in further maths
0
16 years ago
#2
(Original post by matouwah)
a) Defferenciate x^(x^2) with respect to x

is this **** done in further maths

what module is it?
0
16 years ago
#3
(Original post by matouwah)
a) Defferenciate x^(x^2) with respect to x

is this **** done in further maths
let y=x^2

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

=> y'= x^(x^2)[2xlnx+x]
0
#4
(Original post by ShOcKzZ)
what module is it?
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...
0
16 years ago
#5
(Original post by 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)
0
#6
(Original post by 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...
0
16 years ago
#7
(Original post by 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))
= ...
0
#8
(Original post by 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...
0
16 years ago
#9
take logs.
0
16 years ago
#10
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.
0
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