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!
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!
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.
$\displaystyle x(x+1)^{-4}=(x+1)^{-4}-4x(x+1)^{-5}=(x+1)^{-5}((x+1)-4x)=\frac{-3x+1}{(x+1)^5}$
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
$\displaystyle 1 -3x / (x + 1)^5 = 1 - \frac{3x}{(x + 1)^5}$
Hi Plato, there is a small (probably typing error) in your answer for f'.
Salahuddin
Maths online