Apply the generalized product rule.

$\displaystyle x^x$

Online Derivative Calculator ? Shows All Steps!

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

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

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.

Re: Apply the generalized product rule.

Quote:

Originally Posted by

**sepoto** $\displaystyle x^x$

Online Derivative Calculator ? Shows All Steps!
My derivative calculator says to apply the generalized product rule so that:

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

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 x^{x} = e^{x ln(x)}

and that d/dx (e^{f(x)}) = e^{f(x)}*df/dx(x)

that should be all you need to compute your desired derivative.

Re: Apply the generalized product rule.

$\displaystyle e^{xln(x)}$

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 $\displaystyle x^x$?

Re: Apply the generalized product rule.

Quote:

Originally Posted by

**sepoto** $\displaystyle e^{xln(x)}$

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 $\displaystyle x^x$?

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

-Dan

Re: Apply the generalized product rule.

Thank you for clearing that up. So then:

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

That's interesting so I think the above looks a little bit like the chain rule but I'm not sure yet about that.

Re: Apply the generalized product rule.

It **is** the chain rule. letting $\displaystyle y= x ln(x)= x^x$, $\displaystyle e^{xln(x)}= e^y$.

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

dy/dx= (x ln(x))' and $\displaystyle e^y= e^{x ln(x)}$ so $\displaystyle (e^{x ln(x)})'= (x ln(x))' e^{x ln(x)}$

Re: Apply the generalized product rule.

So the derivative of $\displaystyle e^x$ is $\displaystyle e^x$

I am seeing the chain rule as:

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

Is that above not correct?

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

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

The:

$\displaystyle e^{x*ln(x)}$

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

Thanks for any responses...

Re: Apply the generalized product rule.

Quote:

Originally Posted by

**sepoto** So the derivative of $\displaystyle e^x$ is $\displaystyle e^x$

I am seeing the chain rule as:

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

Is that above not correct?

No, this isn't correct

(e^{f(x)})' = f'(x) e^{f(x)}

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

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

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

Thanks for any responses....

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

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

In the case $\displaystyle f'(x)*e^{f(x)}$ my question is that $\displaystyle e^{f(x)}$ does not appear to be a derivative in this case. So $\displaystyle f'(x)$ clearly is a derivative but $\displaystyle e^{f(x)}$ 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(u^{2}) where u = 3x+1

d/dx(u^{2}) = d/du(u^{2}) du/dx = 2u * 3 = 2(3x+1)*3 = 6(3x+1)

Apply this thinking to your exponential example

d/dx (e^{f(x)})

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

d/dx(e^{u}) = d/du(e^{u}) du/dx = e^{u} f'(x) = e^{f(x)} f'(x)