How to solve for the derivative of (1/X+1) +X +1
\[\frac{ 1 }{ X+1 } +X +1\]
you know how to differentiate x+1?
for the first term you can use quotient rule for the second term you can use power rule for the third term you can use constant rule
is the constant rule where it stays the same? so there are three terms right, i don't understand why the second term uses power wouldn't it be the f(x) +or- g(x)?
HI!!
\[\frac{d}{dx}(constant)=0 \]
\[\frac{d}{dx}(\frac{1}{x+1}+x+1) \\ =\frac{d}{dx}(\frac{1}{x+1})+\frac{d}{dx}(x)+\frac{d}{dx}(1)\]
you could actually write 1/(x+1) as (x+1)^(-1) and forget about quotient if you prefer
you could look at the 2nd and 3rd as one ... if you want and ask yourself what is the slope of the line y=x+1
okay i understand now thank you i think i might skip the quotient and implement a negative power like you said. thanks
did you want to write what you have so I can check it?
sure thanks \[\frac{ 1}{ x+1 } +x+1 \]
\[(x+1)^{-1}\]
i meant \[(x+1)^{-1}+1+1\]
\[\frac{1}{x+1}+x+1=(x+1)^{-1}+x+1\]
then \[-1(x+1)^{-2} +1+1\]
whait i thought the x became a 1 because of the power rule?
\[\frac{d}{dx}(\frac{1}{x+1}+x+1) \\ \frac{d}{dx}(x+1)^{-1}+\frac{d}{dx}x+\frac{d}{dx}1 \\ -1(x+1)^{-2}+1+0\]
ohhh i seee
dx/dx is 1 but d(constant)/dx=0
the answer is supposed to be \[\frac{ x ^{2}+2x }{ (x+1)^{2} }\] i can see how the bottom works out but no the top
\[\frac{d}{dx}(\frac{1}{x+1}+x+1) \\ \frac{d}{dx}(\frac{1}{x+1})+\frac{d}{dx}(x) +\frac{d}{dx}(1) \\ \frac{d}{dx}(x+1)^{-1} +1+0 \\ -1(x+1)^{-2}+1 \\ -\frac{1}{(x+1)^2}+1\] try combining the fractions
\[\frac{-1}{(x+1)^2}+1=...\]
remember you can write 1 as (x+1)^2/(x+1)^2 this will allow you have the same denominator so your can write the two fractions as one
you can try wolframalpha.com they also show great explanations
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