Question

differentiate the function:
f(x)= x^4(1-(2/x+1))

Answer 1

i just need help figuring out the f and g primes

Answer 2

for the f prime i have 4x^3

Answer 3

do you know tha chain rule as a generalization of diferenciating almost all kind of functions?

Answer 4

but the g prime has me a little confused

Answer 5

\[x^4(1-\frac{2}{x+1})\]?

Answer 6

yes, like that satellite. also i'm familiar with the chain rule. i'm just having trouble with the g prime here

Answer 7

my guess for the g prime was, ((-2/x+1)+1) = -2^-1x+1 + 1= 2x^-x

Answer 8

the derivative of 1 is zero, and the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\) so the derivative of
\[1-\frac{2}{x+1}\] is
\[\frac{2}{(x+1)^2}\]

Answer 9

oh wow, i've never seen it like that. hmm

Answer 10

you lost me on your guess, but it looks like you are trying to use the power rule
while this rule always works, there are a couple derivatives you should memorize, because they come up again and again
\[f(x)=\frac{1}{x}\] is a rather common function (take the reciprocal) and its derivative is
\[f'(x)=-\frac{1}{x^2}\]

Answer 11

you should also try to remember that the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\) because the square root in another really really common function

Answer 12

if you can remember these, and the chain rule, then you can see that the derivative of
\[\frac{1}{f(x)}\] is
\[-\frac{f'(x)}{f^2(x)}\] and the derivative of
\[\sqrt{f(x)}\] is \[\frac{f'(x)}{2\sqrt{f(x)}}\] save you a ton of time on exams, when everyone else it trying to recall that \(\sqrt{x}=x^{\frac{1}{2}}\) etc

Answer 13

i'll have to study those a bit more but i understand what you mean. thank you very much

Answer 14

yw
btw i really just mean "memorize" like knowing that \(7\times 8=56\) right off the top of your head
good luck!