What is the Derivative of xlnx and Why is it 1 + lnx?

[Shay10825]

Hi everyone!

(d/dx) [xlnx]

Why is the answer 1+lnx and not 1??

~Thanks

 

[Curious3141]

Do you know the product rule ?

 

[recon]

$$\frac{d\ln x}{dx} = \frac{1}{x}$$, in case you didn't know.

 

[courtrigrad]

$$f'(x) = x * (1/x) + ln x * 1 = 1 + ln x$$
$$d(uv) = vdu + udv$$

 

[Shay10825]

Wow I'm really stupid! I don't know how I did not see that. Thanks

 

[Shay10825]

At first I did:
x(1/x) which gave me 1 but why does this not work:

xlnx
ln (x^x)
[x(x^x-1)]/(x^x) ?

I know the second way is not how you would usually do it but why does it not work??

 

[recon]

What you've just written is very confusing, do you mind trying to make it a bit more clear?

 

[Shay10825]

I'm sorry.

http://img108.exs.cx/my.php?loc=img108&image=equation8ag.png

 

[Curious3141]

At first I did:
x(1/x) which gave me 1 but why does this not work:

xlnx
ln (x^x)
[x(x^x-1)]/(x^x) ?

I know the second way is not how you would usually do it but why does it not work??

$$\frac{d}{dx}(x^n) = nx^{n-1}$$ only works when n is a constant. Can you see your error now ?

 

[courtrigrad]

the derivative of x^x is not $$x( x^x - 1)$$. It's $$x^x((ln(x) + 1))$$

 

[Shay10825]

Yeah I see my error. Thanks.

 

[Shay10825]

the derivative of x^x is not $$x( x^x - 1)$$. It's $$x^x((ln(x) + 1))$$

Is there a rule for this or something?

 

[courtrigrad]

$$x^x(ln(x) +1 )$$

$$f(x) = a^x$$ then $$f'(x) = a^x(ln(a))$$

 

[Shay10825]

Thanks

 

[Curious3141]

Is there a rule for this or something?

You don't need a rule.

$$x^x = e^{x\ln x}$$. Can you see how to differentiate it now ?

The other less direct but "easier to see" way is to use implicit differentiation.

I once "just used" the derivative of $x^x$ in exam and got docked a couple of points spoiling an otherwise perfect score. The teacher refused to believe I just did it in my head.

 

[Shay10825]

You don't need a rule.

$$x^x = e^{x\ln x}$$. Can you see how to differentiate it now ?

The other less direct but "easier to see" way is to use implicit differentiation.

I once "just used" the derivative of $x^x$ in exam and got docked a couple of points spoiling an otherwise perfect score. The teacher refused to believe I just did it in my head.

I'm sorry but you just lost me. How did you get $$x^x = e^{x\ln x}$$?

 

[Curious3141]

I'm sorry but you just lost me. How did you get $$x^x = e^{x\ln x}$$?

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

That's actually the easy part.

 

[Shay10825]

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

Do you have to memorize this or something?

 

[Curious3141]

Do you have to memorize this or something?

No, isn't it obvious ? I'm just using $a = e^{\ln a}$ and ${(a^b)}^c = a^{bc}$

 

[courtrigrad]

$$x^x = (e^{\ln x})^x = e^{x\ln x}$$ there is also a rule which states this. But as Curious said, its pretty obvious.

 

[Shay10825]

Ok I see it now. I never knew $a = e^{\ln a}$. My teacher just gave us a bunch or rules to memorize. Is there a way I can do these kind of problems without the rules? or do I have to memorize them?

 

[Yapper]

How does changing it into e^xlnx make it easier?

 

[Curious3141]

How does changing it into e^xlnx make it easier?

Because it is easy to differentiate a form $e^{f(x)}$

$$\frac{d}{dx}(e^{f(x)}) = f'(x)e^{f(x)}$$

Now use that to find the differential of $x^x$.

 

[Yapper]

but f(prime) of x is the derivative orignal equation xlnx so how does that help?

 

[Curious3141]

but f(prime) of x is the derivative orignal equation xlnx so how does that help?

$$y = x^x = e^{x\ln x}$$

Let $f(x) = x\ln x$

$$f'(x) = 1 + \ln x$$

$$\frac{dy}{dx} = f'(x)e^{f(x)} = (1 + \ln x)e^{x\ln x} = x^x(1 + \ln x)$$

Happy ?