Problem in logarithmic differentiation

[jw1180]

[MATH][/MATH]Hello,

I am working through some logarithmic differentiation problems. The following example was presented:

Differentiate y = x^x(1/2)

Using the logarithmic differentiation and the product rule, the example provides the answer:

y' = x^x(1/2)[(2 + ln x) / 2x^(1/2)]

Why can't I do the following instead?

y = x^x(1/2)

log_xy = log_xx^x(1/2) = x^(1/2)

(1/(y ln x))y' = (1/2)x^-(1/2)

y' = (y ln x)/(2x^(1/2))

(substituting the first equation for y):

y' = x^x(1/2)(ln x/2x^(1/2))

which as far as I can tell does not agree with the above method. Any help is appreciated!

(ref: Stewart, J. Calculus, Early Transcendentals (5th ed). Belmont, CA, Brooks/Cole, 2003. Page 247.)

 

[JeffM]

I am guessing that the problem specifies x > 0, or else I am going to be having problems at the very outset.

Now looking at your method, x is a variable. Is log_1(y) even meaningful?

Ignoring that issue, which may be fatal on its own, I guess you went

[MATH]log_x(y) = x^{1/2} \implies \dfrac{ln(y)}{ln(x)} = x^{1/2}[/MATH].

But is that change of base formula valid if y is a function of a variable base. I have not seen a proof. Do you have one? I am not saying it is wrong, but you need a proof.

[MATH]\text {Let } u = ln(x), \ v = ln(x^{\sqrt{x}}), \text { and } w = \dfrac{u}{v} \implies w’ = \dfrac{vu’ - uv’}{v^2}?.[/MATH]
Just what does that equal?

You seem to have been totally casual about the difference between a constant and a variable.

 

[Steven G]

log_xy = log_xx^x1/2 = x^(1/2)

The way you computed the derivative of log_xy was wrong! You treated x as a constant.

The derivative of log_xy = the derivative of lny/lnx. This requires the quotient rule!

Please try again!

EDIT: this is what Jeff said!

 

[pka]

I am working through some logarithmic differentiation problems.
Differentiate y = x^x(1/2)
Using the logarithmic differentiation and the product rule, the example provides the answer:
y' = x^x(1/2)[(2 + ln x) / 2x^(1/2)]
(ref: Stewart, J. Calculus, Early Transcendentals (5th ed). Belmont, CA, Brooks/Cole, 2003. Page 247.)

I honestly don't know what you are asking.
But if \(y=x^{\sqrt{x}}\) then \(\log(y)=\sqrt{x}\log(x)\).
Then \(\dfrac{y^{\prime}}{y}=\dfrac{\log(x)}{2\sqrt{x}}+\dfrac{\sqrt{x}}{x}\) or
\(y^{\prime}=x^{\sqrt{x}}\left(\dfrac{\log(x)}{2\sqrt{x}}+\dfrac{\sqrt{x}}{x}\right)=\dfrac{1}{2}x^{\sqrt{x}-\frac{1}{2}}(\log(x)+2)\) SEE THIS

 

[jw1180]

Thank you all for the detailed replies. I understand now...as Jomo and JeffM pointed out, I mistakenly treated x as a constant when I computed log_x y.

The recommended method in the textbook is to use

ln y = ln x^x(1/2) = x^(1/2) ln x

and proceed from there, as detailed by user pka.

Thanks again!

 

[Steven G]

You computed log_xy perfectly fine. The derivative of log_xy was where the error.