Derivative of: f(x) = cosx^xlnx

[Math_Junkie]

Any help? I'm completely lost with all the steps involved..

Thanks.

 

[daon]

This should start you off:

\(\displaystyle [(cosx^x)(lnx)]' = [cosx^x]'(lnx) + (cosx^x)[lnx]'\)

 

[Math_Junkie]

The 'xlnx' is all one exponent, so maybe I should have written it like this: f(x) = cosx^(xlnx).

 

[daon]

In that case:

\(\displaystyle y = f(x)^{g(x)}\)

\(\displaystyle ln(y) = g(x)ln[f(x)]\)

Which implies:

\(\displaystyle \frac{1}{y}\frac{dy}{dx} = [g(x)ln(f(x))]'\)

And hence:

\(\displaystyle \frac{dy}{dx} = f(x)^{g(x)}\left [g'(x)ln(f(x)) + \frac{g(x)f'(x)}{f(x)} \right ]\)

 

[mmm4444bot]

… maybe I should have written it like this: f(x) = cosx^(xlnx)

This is still ambiguous. Do you realize that cosine is a function? So is the natural logarithm. Use function notation, when working with functions.

f(x) = cos(x)^(x ln[x])

OR

f(x) = cos(x^[x ln(x)])

:?:

 

[BigGlenntheHeavy]

My way:

\(\displaystyle f(x) \ = \ y \ = \ [cos(x)]^{xln|x|}, \ ln|y| \ = \ xln|x|ln|cos(x)|.\)

\(\displaystyle Ergo, \ y \ = \ e^{xln|x|ln|cos(x)|}\)

\(\displaystyle Then \ y \ ' \ = \ \bigg[(1)ln|x|ln|cos(x)|+x(1/x)ln|cos(x)|+xln|x|\bigg[\frac{-sin(x)}{cos(x)}\bigg]\bigg][cos(x)]^{xln|x|}, \ e^{xln|x|ln|cos(x)| \ =\)\(\displaystyle \ [cos(x)]^{xln|x|}\)

\(\displaystyle Hence \ y \ ' \ = \ [ln|x|ln|cos(x)|+ln|cos(x)|-xln|x|tan(x)][cos(x)]^{xln|x|}.\)

 

[Deleted member 4993]

Oh then -- I have to show mine!!!

y = [cos(x)][sup:2j2pwsvw]x*ln(x)[/sup:2j2pwsvw]

ln(y) = x * ln(x) * ln[cos(x)]

y'/y = ln(x) * ln[cos(x)] + ln[cos(x)] - x * ln(x) * tan(x)

y' = {ln(x) * ln[cos(x)] + ln[cos(x)] - x * ln(x) * tan(x)} * [cos(x)][sup:2j2pwsvw]x*ln(x)[/sup:2j2pwsvw]

.