Differentiate y = x^ln(x)?

Question

Okay, so I used logarithmic differentiation but my answer is different from the teacher's

So I did this

y = x^ln(x)
ln y = ln[x^ln(x)]
ln y = ln(x) * ln(x)
ln y = ln(x)^2
1/y * dy/dx = 2ln(x)  [Power Rule...?]
dy/dx = y * 2ln(x)
dy/dx = [x^ln(x)] * [2ln(x)]

My teacher's answer is
dy/dx = [x^ln(x)] * [2/x * ln(x)]

I can't seem to figure out why he divided the 2 by x. Perhaps I made an error somewhere? Could someone please help me, thanks.

Geronimo · Accepted Answer

ln(x)  •  ln(x)  =   ln ² (x)  =  (  ln(x)  ) ²

The chain-rule derivative:  d{  ln ² (x)  } ⁄ dx

d[  ln(x)  ] ² ⁄ dx ———>  2 • [  ln(x)  ] ¹ • d{ ln(x) }  ⁄ dx

d[  ln(x)  ] ² ⁄ dx ———>  2 • [  ln(x)  ] ¹ • (1 ⁄ x)

d[  ln(x)  ] ² ⁄ dx ———>  2 • ln(x)  ⁄ x

Anonymous · Answer

y = x^ln(x)
ln y = ln(x^lnx)
ln y = [ln (x)]^2
Differentiate
(1/y)dy/dx = 2ln(x)/x
dy/dx = y(2ln(x))/x
dy/dx = (x^lnx)(2ln(x)) / x

Edit: I just read your problem.

You made your error while differentiating [ln(x)]^2
You use chain rule here:
[ln(x)]^2
First the exponent, then the natural logarithm.

Hope I helped!

Trending News

Differentiate y = x^ln(x)?

2 Answers