# simple derivative of...

• Oct 26th 2008, 02:47 PM
juanfe_zodiac
simple derivative of...
could someone help me with this: if y= 1/(x+1) find y'...

i tried and i got this equation x^2 +2x +1 is it correct???
• Oct 26th 2008, 03:00 PM
Jhevon
Quote:

Originally Posted by juanfe_zodiac
could someone help me with this: if y= 1/(x+1) find y'...

i tried and i got this equation x^2 +2x +1 is it correct???

no

write as $y = (x + 1)^{-1}$

now use the chain rule
• Oct 26th 2008, 03:40 PM
juanfe_zodiac
simple derivative of...
i still dont getting it...could u explain me...what chain is that? how do i know that the new derivative is correct?
• Oct 26th 2008, 03:47 PM
RedBarchetta
$
\begin{gathered}
y = \frac{1}
{{x + 1}} \hfill \\
y' = ? \hfill \\
\end{gathered}
$

Use the quotient rule:

$
\left( {\frac{u}
{v}} \right)' = \frac{{u'v - uv'}}
{{v^2 }}
$

Where u=1 and v=x+1.....
• Oct 26th 2008, 05:16 PM
Jhevon
Quote:

Originally Posted by juanfe_zodiac
i still dont getting it...could u explain me...what chain is that? how do i know that the new derivative is correct?

the chain rule says: $\frac d{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

basically, to find the derivative of a composite function, differentiate the outer function (leaving the inner function intact) then multiply by the derivative of the inside function

$y = (x + 1)^{-1}$

$\Rightarrow y' = -(x + 1)^{-2}(1) = \frac {-1}{(x + 1)^2}$

as for knowing the derivative is correct, look up the answer in your text :D...or check it with integration, but i don't suppose you have started integration as yet