Question

f(x)= 4x/x^2-25
need f' and f".....I tried and failed,thanks.

Answer 1

So far I have f'=4x^-2-4/25, f"=-8/x^3

Answer 2

\(\Huge f(x)= \frac{4x}{x^2}-25\)
or \(\Huge f(x)= \frac{4x}{x^2-25}\)

Answer 3

second one. should have used ( )

Answer 4

ok. did you just learn the quotient rule?

Answer 5

yeah. what I got above is wrong

Answer 6

should I not have simplified?

Answer 7

so then take
u=4x
u'=4
v=x^2-25
v'=2x
\(\Huge \frac{4(x^2-25)-8x^2}{(x^2-25)^2}=\frac{-12x^2-100}{(x^2-25)^2}\)

Answer 8

wouldnt it be -4x^2-100?

Answer 9

yes.

Answer 10

so no simplifying past this point?

Answer 11

\(\Huge \frac{4(x^2-25)-8x^2}{(x^2-25)^2}=\frac{-4x^2-100}{(x^2-25)^2}=-4\frac{x^2+25}{(x^2-25)^2}\)
that's probably the most you can do.

Answer 12

I know it's cut off but I just factored out the 4 fo rthe last part. I would suggest you either expand the (x^2-25)^2 or factor it out completely into (x-5)^2*(x+5)^2

Answer 13

x^2+25 does not factor into the reals.

Answer 14

so what we are left with for f'= -4 (x^2+25)/(x-5)^2*(x+5)^2

Answer 15

yeah, make sure you put parenthesis f'= -4 (x^2+25)/((x-5)^2*(x+5)^2)

Answer 16

for f'' it's just more grinding. Factor out the -4, and then start with u and v
u=x^2+25
u'=2x
v=(x^2-25)^2
v'=2(x^2-25)(2x) (chain rule)
and just a bunch of algebra next