Find the derivative of this function

[Helly123]

Homework Statement

$$y = x.\log_e {\sqrt{x}}$$

Homework Equations

f(x) = g(x) h(x)
f ' (x) = g ' (x) . h (x) - h ' (x) . g(x)

The Attempt at a Solution

$$y = x .\log_e {\sqrt {x}}$$
$$y '(x) = 1.ln \sqrt{x} + \frac{1}{2} $$

the right answer is
$$ y ' = \log_{10} {\sqrt{x}} + \frac{1}{2} $$

$$ Is \ \log_e {\sqrt{x}} = \log_{10} {\sqrt{x}} \ ? $$

Second Try:
$$ y = x \log_e {\sqrt{x}} $$
$$ y = \log_e {x^{x/2}} $$
$$y ' = \frac{1}{x^{x/2}} . \frac{x}{2} . x^{(x-2)/2}$$
$$ y ' = \frac{1}{2}$$

why different result?

 

[PeroK]

What is the derivative of $\ln \sqrt{x}$?

Hint: Chain rule!

 

[scottdave]

log_e is the natural log. I don't know where they got log₁₀.
In your second attempt, you cannot take derivative of x^x or x^x/2 like that.

If you use Wolframalpha and some other solvers, the natural log is often denoted by just log() with no subscript. Perhaps that's why you think it should be base 10? But your first attempt is not quite right.

Try your product rule again, taking care to not drop any factors of 1/2 along the way.

 

[hilbert2]

Also note that $\log \sqrt{x} = \log x^{1/2} = \frac{1}{2}\log x$, regardless of the base of the logarithm.

 

[PeroK]

Also note that $\log \sqrt{x} = \log x^{1/2} = \frac{1}{2}\log x$, regardless of the base of the logarithm.

That's definitely the simplest approach but perhaps the OP needs some practice with the chain rule.

 

[hilbert2]

That's definitely the simplest approach but perhaps the OP needs some practice with the chain rule.

Good point.

 

[Helly123]

What is the derivative of $\ln \sqrt{x}$?

Hint: Chain rule!

derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{2}}$

 

[Dick]

derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{2}}$

Show how you derived that. Where did the $\sqrt{2}$ come from? Did you mean that to be $\sqrt{x}$?

 

[Helly123]

Show how you derived that. Where did the $\sqrt{2}$ come from? Did you mean that to be $\sqrt{x}$?

derivative of $ln \ u = \frac{1}{u} . \frac{d(u)}{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{ d ( \sqrt{x} ) \ }{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{2}}$

 

[Dick]

derivative of $ln \ u = \frac{1}{u} . \frac{d(u)}{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{ d ( \sqrt{x} ) \ }{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{2}}$

Hmm. Try $\frac{ d ( \sqrt{x} ) \ }{dx}$ again. Write it as $\frac{ d ( x^{\frac{1}{2}}) \ }{dx}$.

 

[Helly123]

Hmm. Try $\frac{ d ( \sqrt{x} ) \ }{dx}$ again. Write it as $\frac{ d ( x^{\frac{1}{2}}) \ }{dx}$.

$\frac{ d ( x^{\frac{1}{2}}) \ }{dx}$
$= \frac{1}{2}. x^{(\frac{1}{2} - 1)}$.
$\frac{1}{2}.\frac{1}{\sqrt{2}}$

 

[Dick]

$\frac{ d ( x^{\frac{1}{2}}) \ }{dx}$
$= \frac{1}{2}. x^{(\frac{1}{2} - 1)}$.
$\frac{1}{2}.\frac{1}{\sqrt{2}}$

$x^{-\frac{1}{2}}$ is not equal to $\frac{1}{\sqrt{2}}$! Look at it carefully... You have to mean $\frac{1}{\sqrt{x}}$. They aren't the same thing.

 

[Helly123]

$x^{-\frac{1}{2}}$ is not equal to $\frac{1}{\sqrt{2}}$! Look at it carefully... You have to mean $\frac{1}{\sqrt{x}}$. They aren't the same thing.

i'm sorry. yes I meant $\frac{1}{\sqrt{x}}$.

 

[Helly123]

derivative of $ln \ u = \frac{1}{u} . \frac{d(u)}{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{ d ( \sqrt{x} ) \ }{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{x}}$

 

[Dick]

derivative of $ln \ u = \frac{1}{u} . \frac{d(u)}{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{ d ( \sqrt{x} ) \ }{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{x}}$

That's it. And that agrees with what you would get by differentiating $\frac{1}{2} \ln{x}$, yes?

 

[Mark44]

derivative of $ln \ u = \frac{1}{u} . \frac{d(u)}{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{ d ( \sqrt{x} ) \ }{d(x)}$
derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{x}}$

Do you see that the last expression can be simplified, and should be?

 

[Helly123]

That's it. And that agrees with what you would get by differentiating $\frac{1}{2} \ln{x}$, yes?

yes, thanks

 

[Helly123]

Do you see that the last expression can be simplified, and should be?

derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{x}}$
derivative of $\ln \sqrt{x} = \frac{1}{2{x}}$

 

[Mark44]

derivative of $\ln \sqrt{x} = \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{x}}$
derivative of $\ln \sqrt{x} = \frac{1}{2{x}}$

Yes, that's what I meant.