# Apply the generalized product rule.

• Nov 26th 2013, 07:57 PM
sepoto
Apply the generalized product rule.
$x^x$

Online Derivative Calculator ? Shows All Steps!

My derivative calculator says to apply the generalized product rule so that:

$x^x*\frac{d}{dx}(ln(x)*x)$

At first I thought it looked kind of like the chain rule however x^x is not the derivative of x^x. The answer output by the derivative calculator matches my solutions manual but I don't understand the steps I think.

Thanks for any responses...

P.S. I the calculator just means the product rule then I am still confused about ln(x) and how ln(x) became a part of the equation.
• Nov 26th 2013, 08:23 PM
romsek
Re: Apply the generalized product rule.
Quote:

Originally Posted by sepoto
$x^x$

Online Derivative Calculator ? Shows All Steps!

My derivative calculator says to apply the generalized product rule so that:

$x^x*\frac{d}{dx}(ln(x)*x)$

At first I thought it looked kind of like the chain rule however x^x is not the derivative of x^x. The answer output by the derivative calculator matches my solutions manual but I don't understand the steps I think.

Thanks for any responses...

P.S. I the calculator just means the product rule then I am still confused about ln(x) and how ln(x) became a part of the equation.

note that xx = ex ln(x)

and that d/dx (ef(x)) = ef(x)*df/dx(x)

that should be all you need to compute your desired derivative.
• Nov 27th 2013, 10:51 AM
sepoto
Re: Apply the generalized product rule.
$e^{xln(x)}$

I am wondering how the equation above became equivalent to [tex]x^x[tex]. Also which x in the equation with e came from the exponent in the original $x^x$?
• Nov 27th 2013, 11:05 AM
topsquark
Re: Apply the generalized product rule.
Quote:

Originally Posted by sepoto
$e^{xln(x)}$

I am wondering how the equation above became equivalent to [tex]x^x[tex]. Also which x in the equation with e came from the exponent in the original $x^x$?

$e^{x \cdot ln(x)} = \left ( e^{ln(x)} \right ) ^x = x^x$

-Dan
• Nov 27th 2013, 11:24 AM
sepoto
Re: Apply the generalized product rule.
Thank you for clearing that up. So then:

$(e^{xln(x)})'=(xlnx)'e^{xln(x)}$????

That's interesting so I think the above looks a little bit like the chain rule but I'm not sure yet about that.
• Nov 27th 2013, 11:34 AM
HallsofIvy
Re: Apply the generalized product rule.
It is the chain rule. letting $y= x ln(x)= x^x$, $e^{xln(x)}= e^y$.

$\frac{de^y}{dx}= \frac{de^y}{dy}\frac{dy}{dx}= \frac{dy}{dx}e^y$

dy/dx= (x ln(x))' and $e^y= e^{x ln(x)}$ so $(e^{x ln(x)})'= (x ln(x))' e^{x ln(x)}$
• Nov 27th 2013, 12:51 PM
sepoto
Re: Apply the generalized product rule.
So the derivative of $e^x$ is $e^x$

I am seeing the chain rule as:

$(e^{x*ln(x)})'*(x*ln(x))'$

Is that above not correct?

I am trying to see how that might be similar to:

$(x*ln(x))'e^{x*ln(x)}$

The:
$e^{x*ln(x)}$

is confusing because I don't see that above as a derivative.

Thanks for any responses...
• Nov 27th 2013, 01:00 PM
romsek
Re: Apply the generalized product rule.
Quote:

Originally Posted by sepoto
So the derivative of $e^x$ is $e^x$

I am seeing the chain rule as:

$(e^{x*ln(x)})'*(x*ln(x))'$

Is that above not correct?

No, this isn't correct

(ef(x))' = f'(x) ef(x)
• Nov 27th 2013, 01:33 PM
sepoto
Re: Apply the generalized product rule.
Thank you for clearing that up. I believe that is true but I am still confused about something. From what I know about the chain rule it is the derivative of the whole part multiplied by the derivative of the inside.

Example:
https://www.math.ucdavis.edu/~kouba/....html#SOLUTION1

$((3x+1)^2)'*(3x+1)'=f'$

In the case $f'(x)*e^{f(x)}$ my question is that $e^{f(x)}$ does not appear to be a derivative in this case. So $f'(x)$ clearly is a derivative but $e^{f(x)}$ does not look like one. Is there something I don't understand maybe about the chain rule?

Thanks for any responses....
• Nov 27th 2013, 03:19 PM
romsek
Re: Apply the generalized product rule.
Quote:

Originally Posted by sepoto
Thank you for clearing that up. I believe that is true but I am still confused about something. From what I know about the chain rule it is the derivative of the whole part multiplied by the derivative of the inside.

Example:
https://www.math.ucdavis.edu/~kouba/....html#SOLUTION1

$((3x+1)^2)'*(3x+1)'=f'$

In the case $f'(x)*e^{f(x)}$ my question is that $e^{f(x)}$ does not appear to be a derivative in this case. So $f'(x)$ clearly is a derivative but $e^{f(x)}$ does not look like one. Is there something I don't understand maybe about the chain rule?

Thanks for any responses....

it might be clearer if you thought of the example about as follows.

d/dx((3x+1)2) = d/dx(u2) where u = 3x+1

d/dx(u2) = d/du(u2) du/dx = 2u * 3 = 2(3x+1)*3 = 6(3x+1)

Apply this thinking to your exponential example

d/dx (ef(x))

let u = f(x) du/dx = f'(x)

d/dx(eu) = d/du(eu) du/dx = eu f'(x) = ef(x) f'(x)