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Differentiation question - using logs

Ok, second question in 30 minutes.

This times it a differentiation question.

"Given that

y = x^x

x > 0

y > 0

, by taking logarithms show that:

\frac{dy}{dx} = x^x(1 + \ln x)

Now I genuinely have no idea how to do this so could somebody please talk me through it.

Thanks.

Reply 1

technicolour

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

Does this help at all? You could use a combination of chain rule and product rule to figure it out from here I think.

Edit: It works.

Reply 2

jayshah31

\ln y = x \ln x

Implicitly differentiate the LHS, and use product rule on the RHS.

Reply 3

nuodai

I was sad enough to make a thread about this a while back.

Reply 4

Prokaryotic_crap

technicolour

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

thats impressive! teach me how you got that. never seen it before.

Reply 5

jayshah31

nuodai

I was sad enough to make a thread about this a while back.

Reply 6

nuodai

Prokaryotic_crap

thats impressive! teach me how you got that. never seen it before.

You know from C2 that

k^{\log_k a} = a

, so it's basically the same thing applied here;

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

... you can verify it, because

\ln {a^b} = b\ln a \Rightarrow \ln x^x = x\ln x

and

e^{\ln k} = k \Rightarrow e^{x\ln x} = x\ln x

Reply 7

Prokaryotic_crap

nuodai

You know from C2 that

k^{\log_k a} = a

, so it's basically the same thing applied here;

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

... you can verify it, because

\ln {a^b} = b\ln a \Rightarrow \ln x^x = x\ln x

and

e^{\ln k} = k \Rightarrow e^{x\ln x} = x\ln x

ah yes, got it thanks

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Differentiation question - using logs

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