Question

Calculate f'(3) to 3 significant figures where
f(t) = (4 t^2 + 3 t - 2 )^{-8}

Answer 1

i got -5.575356455*10^(-07) but it says i'm wrong

Answer 2

So f(t)=(4 t^2 + 3 t - 2 )^{-8}
Did you apply the chain and power rule.

f(t)=[g(t)]^n

f'(t)=n[g(t)]^(n-1)*g'(t)

Answer 3

Also it kinda looks like you added 1 instead of subtracted 1 in the exponent area.

Answer 4

yeah i forgot to subtract one for 3t, the new answer i got is -6.195008551*10^(-08) but i don't think it's right

Answer 5

So can I see what you got for f'?

Answer 6

-16*4^-8*t^-17+-8*3^-8*t^-9

Answer 7

\[ (4 t^2 + 3 t - 2 )^{-8}\]
using the rule
\[ \frac{d}{dx} x^n= n x^{n-1} dx \]
\[ -8 (4 t^2 + 3 t - 2 )^{-8-1} d (4 t^2 + 3 t - 2 )\]
you now have to take the derivative of the stuff in the parens
\[ -8 (4 t^2 + 3 t - 2 )^{-8-1} (8 t^1 + 3 )\]

Answer 8

\[ -8 (4 t^2 + 3 t - 2 )^{-9} (8 t^1 + 3 ) \]

replace t with 3

Answer 9

will it be -4.297715384*10^(-13)

Answer 10

yes, but they want it rounded to 3 significant digits

Answer 11

alright thanks

Answer 12

I hope you got
-4.30 * 10^(-13)

Answer 13

yes. another question.. for the derivative for sin^2(x), i got 2 sin (x) cos(x), but when i try to find f'(4) i got 0.139173101 but the system marks me wrong @phi

Answer 14

I would guess that x is in radians. make sure your calculator is in *radian mode* and not degree mode

Answer 15

that helped.. thanks alot