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Mathematics 9 Online
OpenStudy (johan14th):

How to solve for the derivative of (1/X+1) +X +1

OpenStudy (johan14th):

\[\frac{ 1 }{ X+1 } +X +1\]

OpenStudy (freckles):

you know how to differentiate x+1?

OpenStudy (freckles):

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

OpenStudy (johan14th):

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)?

OpenStudy (misty1212):

HI!!

OpenStudy (freckles):

\[\frac{d}{dx}(constant)=0 \]

OpenStudy (freckles):

\[\frac{d}{dx}(\frac{1}{x+1}+x+1) \\ =\frac{d}{dx}(\frac{1}{x+1})+\frac{d}{dx}(x)+\frac{d}{dx}(1)\]

OpenStudy (freckles):

you could actually write 1/(x+1) as (x+1)^(-1) and forget about quotient if you prefer

OpenStudy (freckles):

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

OpenStudy (johan14th):

okay i understand now thank you i think i might skip the quotient and implement a negative power like you said. thanks

OpenStudy (freckles):

did you want to write what you have so I can check it?

OpenStudy (johan14th):

sure thanks \[\frac{ 1}{ x+1 } +x+1 \]

OpenStudy (johan14th):

\[(x+1)^{-1}\]

OpenStudy (johan14th):

i meant \[(x+1)^{-1}+1+1\]

OpenStudy (freckles):

\[\frac{1}{x+1}+x+1=(x+1)^{-1}+x+1\]

OpenStudy (johan14th):

then \[-1(x+1)^{-2} +1+1\]

OpenStudy (johan14th):

whait i thought the x became a 1 because of the power rule?

OpenStudy (freckles):

\[\frac{d}{dx}(\frac{1}{x+1}+x+1) \\ \frac{d}{dx}(x+1)^{-1}+\frac{d}{dx}x+\frac{d}{dx}1 \\ -1(x+1)^{-2}+1+0\]

OpenStudy (johan14th):

ohhh i seee

OpenStudy (freckles):

dx/dx is 1 but d(constant)/dx=0

OpenStudy (johan14th):

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

OpenStudy (freckles):

\[\frac{d}{dx}(\frac{1}{x+1}+x+1) \\ \frac{d}{dx}(\frac{1}{x+1})+\frac{d}{dx}(x) +\frac{d}{dx}(1) \\ \frac{d}{dx}(x+1)^{-1} +1+0 \\ -1(x+1)^{-2}+1 \\ -\frac{1}{(x+1)^2}+1\] try combining the fractions

OpenStudy (freckles):

\[\frac{-1}{(x+1)^2}+1=...\]

OpenStudy (freckles):

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

OpenStudy (anonymous):

you can try wolframalpha.com they also show great explanations

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