derivative of xlnx-x
Separate it into two derivatives and use the product rule :D
well you need to product rule for xln(x) and -x is differentiated in the normal fashion
\[\large \frac{d}{dx}x\ln (x)-\frac{d}{dx}x\]
thats where i get stuck. I have f'(x)=lnx+x(1/x)-1 What's next?
ok... all you need to do is multiply x by 1/x
\[\large \frac{d}{dx}x\ln(x)=(\frac{d}{dx}x)\ln(x)+x(\frac{d}{dx}\ln(x))=\ln(x)+x\frac{1}{x}=\ln(x)\]
The \(\large x(\frac{1}{x})\) part could be simplied as 1, since \(\LARGE x\frac{1}{x}=\frac{x}{x}=1\)
And you'll get \(\large f'(x)=\ln(x)+1-1\), which is simply \(\large f'(x)=\ln(x)\)
***I forgot a 1 here, sorry about that***\[\large \frac{d}{dx}x\ln(x)=(\frac{d}{dx}x)\ln(x)+x(\frac{d}{dx}\ln(x))=\ln(x)+x\frac{1}{x}=\ln(x)+1\]
You're awesome. Nice avatar pic btw
np and thx :)
would the 1 be negative? lnx-1? It was originally xlnx-x
Nope, the derivative of xln(x) is ln(x)+1 and the derivative of x is 1 (ln(x)+1) - (1) Derivative of xlnx Derivative of x
ok. so then the answer is just ln(x) because the +1 and -1 cancel
Yes! :D
sweet. Thanks again!
Welcome :)
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