Help me find the ans!

• idlittle

  6 Jun 09, 23:11

  Find the 1st- order derivative of y = (x²)^(ln x). Thanks 

• Rednano

  6 Jun 09, 23:13

  by just directly give u answer, when will u really learn how to solve similar equations in the future?

• idlittle

  6 Jun 09, 23:18

  I mean the steps for the ans. Cause my ans is not the same as the ans sheet. The ans is 4lnx^( (2lnx) - 1 ).

• deepak.c

  7 Jun 09, 01:42

   

  Not very sure if this is correct.

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

  ln both sides:

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

  ln y = 2 (ln x)^2

  1/y (dy/dx) = 4 (ln x) (1/x)

  dy/dx = (y) (4/x) (ln x)

  dy/dx = (x^2)^(ln x) (4/x) (ln x)

  dy/dx = [4(x^2)^(ln x) * (ln x)] / x

   

  My algebra not powerful enough to simplify to your answer, provided my answer is correct.

   

  Maybe you wait for Mr. Wee or eagle to come help.

   

• deepak.c

  7 Jun 09, 02:34

   

  Strange......

  When I use d/dx a^u = (ln a) a^u (du/dx)

   

  I get,

   

  dy/dx = (ln x) (x^2)^(ln x) (2/x)

  dy/dx = [2(x^2)^(ln x) * (ln x)] / x

   

  The constant 4 became a 2?

   

   

• Forbiddensinner

  7 Jun 09, 13:54

  Originally posted by deepak.c:

   

  Not very sure if this is correct.

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

  ln both sides:

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

  ln y = 2 (ln x)^2

  1/y (dy/dx) = 4 (ln x) (1/x)

  dy/dx = (y) (4/x) (ln x)

  dy/dx = (x^2)^(ln x) (4/x) (ln x)

  dy/dx = [4(x^2)^(ln x) * (ln x)] / x

   

  My algebra not powerful enough to simplify to your answer, provided my answer is correct.

   

  Maybe you wait for Mr. Wee or eagle to come help.

   

  Continuing from the underlined equation,

  dy/dx = 4 (Inx) ( x^-1) ( x^ (2Inx))

  dy/dx = 4 (Inx) (x ^ (2Inx - 1))

  dy/dx = 4 (In (x^(x^(2Inx - 1))))

  It is still different from the answer of 4(In (x^(2Inx -1))) though.

• Lee012lee

  7 Jun 09, 19:43

  The question is fun.

   

  Question

   

  Differentiate y = (x²)^lnx

   

  Steps

   

  (1)    x = e^lnx    eg   e^{ln 3} = 3

   

  (2)    x^lnx = (e^lnx) ^lnx

          

          x^lnx = e ^{(lnx)^2}

   

  (3)   y = (x²)^lnx

   

         y = [x^lnx]²

   

         Substitute x^lnx = e ^{(lnx)^2}

   

         y = [e ^{(lnx)^2}]^{^2}

   

  (4)  Using Chain Rule,

   

         dy/dx = 2 [e ^{(lnx)^2}]^{^1}[e ^{(lnx)^2}] [2 lnx] [1/x]

   

  (5)  Substitute e ^{(lnx)^2} = x^lnx

   

  (6)  dy/dx = 2 [x^lnx]^{^1}[x^lnx] [2 lnx] [1/x]

   

         dy/dx = 2 [x^lnx]^{^1}[x^lnx] [2 lnx] [x]^-1

   

         dy/dx = 4ln x [x^{2lnx – 1}] 

• deepak.c

  7 Jun 09, 20:57

   

  The answer provided by Forbiddensinner is similar. I think TS might have mistyped the answer.

   

  dy/dx = 4 (Inx) (x ^ (2Inx - 1))

   

  dy/dx = 4ln x [x^{2lnx – 1}] 

• idlittle

  7 Jun 09, 21:41

  Thanks a lot.

  Yep, i typed the ans wrongly. lol.