taking the second derivative of a rational function

May 2012
277
1
Toronto
Taking the second derivative of \(\displaystyle x/(x+1)^{4}\)


for the first derivative I got

\(\displaystyle 1-3x/(x+1)^5\)

the second derivative I'm not sure about. Im not even sure I did the first one correctly.


thanks!
 
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TheEmptySet

MHF Hall of Honor
Feb 2008
3,764
2,029
Yuma, AZ, USA
Taking the second derivative of \(\displaystyle x/(x+1)^{4}\)


for the first derivative I got

\(\displaystyle ((x+1)^3-4x)/(p+1)^7\)

the second derivative I'm not sure about. Im not even sure I did the first one correctly.


thanks!


No the first derivative is not correct.

I am not a fan of the quotient rule, so I don't usually use it. Just use negative exponents to rewrite and use the product rule.

\(\displaystyle x(x+1)^{-4}=(x+1)^{-4}-4x(x+1)^{-5}=(x+1)^{-5}((x+1)-4x)=\frac{-3x+1}{(x+1)^5}\)
 

Plato

MHF Helper
Aug 2006
22,508
8,664
Taking the second derivative of \(\displaystyle x/(x+1)^{4}\)
Write it as \(\displaystyle y=x(x+1)^{-4}\)
Then \(\displaystyle y'=(x+1)^{-4}+x[-4(x+1)^{-5}]\)

What is next?
 
Last edited:
Oct 2012
8
2
Lisbeth Salander's apartment
Taking the second derivative of \(\displaystyle x/(x+1)^{4}\)


for the first derivative I got

\(\displaystyle 1-3x/(x+1)^5\)

the second derivative I'm not sure about. Im not even sure I did the first one correctly.


thanks!
Your first derivative is correct, though you should write it properly as
\(\displaystyle f'(x) = \frac{-3x + 1}{(x + 1)^5}\)

So now do your quotient rule again:
\(\displaystyle f''(x) = \frac{(-3)(x + 1)^5 - (-3x + 1)*5(x + 1)^4}{(x + 1)^{10}}\)

-Dan

FYI By order of operations
\(\displaystyle 1 -3x / (x + 1)^5 = 1 - \frac{3x}{(x + 1)^5}\)