What is the derivative of y= (xlnx)/(1+lnx)?

Question

tkhunny · Answer

Well, you have some Product Rule and Quotient Rule in your future.  Let's see your best efforts.

anonymous · Answer

$$
y =\frac{x \cdot ln(x)}{1 + ln(x)}
$$

$$ 
f(x) = x \cdot ln(x) \
f'(x) = \bigg[ \; x \cdot ln(x) \; \bigg]' = (x)' \cdot ln(x) + x \cdot \big[ \; ln(x) \; \big]' =\
= 1 \cdot ln(x) + x \cdot \frac{1}{x} = ln(x) + 1
$$

$$
g(x) = 1 + ln(x) \
g'(x) = \bigg[ \; 1 + ln(x) \; \bigg]' = (1)' + \big[\; ln(x) \; \big]' =0 +\frac{1}{x} = \frac{1}{x} = x^{-1} \
$$

$$
y =\frac{x \cdot ln(x)}{1 + ln(x)} = \frac{f(x)}{g(x)}\
y' = \bigg[\frac{f(x)}{g(x)} \bigg]' = \frac{f'(x)g(x) - f(x)g'(x)}{\big[g(x)\big]^2} = \
= \frac{
(ln(x) + 1)(1 + ln(x)) - x\cdot ln(x) \cdot x^{-1}
}
{ \big[ \; ln(x) + 1 \; \big]^2 }=\
= 1 - \frac{
ln(x)
}
{ \big[ \; ln(x) + 1 \; \big]^2 }
$$
If I got it right..

anonymous · Answer

Seems right: http://www.wolframalpha.com/input/?i=%28x*ln%28x%29%29%2F%28ln%28x%29+%2B+1%29 Since $$ 1 - \frac{ln(x)}{\big[\;ln(x) + 1\;\big]^2 } = \frac{\big[\;ln(x) + 1\;\big]^2 -ln(x)}{\big[\;ln(x) + 1\;\big]^2 } = \ = \frac{ln^2(x) + 2ln(x) + 1 -ln(x)}{\big[\;ln(x) + 1\;\big]^2 } = \ = \frac{ln^2(x) + ln(x) + 1}{\big[\;ln(x) + 1\;\big]^2 } $$ Which is what's in there