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f(x)=x^(x/2) what is f'(x)?
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\[f(x)=x^{x/2}\]
Oh this is a nice one. Do you know logarithmic differentiation?
say y=x^(x/2) ln y =x/2 lnx d(ln y)=d(x ln x)/2 y'/y=1/2 * (ln x+x*1/x) y'/y=1/2*(ln x+1) y'=y/2 * (ln x+1) y'=x^(x/2)*(ln x+1)
Sorry, I shouldn't post the answer right away. But I hope this helps :)
And i missed our 1/2 at the final answer.
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@imron07 you forgot the 1/2 in the final result.
hehe, thanks @gezimbasha
thanks. but how did d(ln y)= y'/y ?
Actually, d(ln x)/dx =1/x For u(x), use chain rule: d(ln u)/dx=1/u * du/dx =u'/u
oh. ok. thanks. :)
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Okay :)
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