Question

Find the derivative

Answer 1

\[\[\huge f(t)= \frac{ t^2-1 }{ t+4 }\]\]

Answer 2

quotient rule!

Answer 3

im doing something wrong. .

Answer 4

What did you get?

Answer 5

i keep getting this : \[\frac{ t^2+8t-1 }{ (t+4)^2 }\]

Answer 6

sorry i mean that should be the answer

Answer 7

but i get 3t^2 as the first term

Answer 8

\[f \prime(x)g(x)-f(x)g \prime(x)\div[g ^{2}] \]

Answer 9

wolfram says the numerator is t^2 + 8t + 1

Answer 10

I got t^2

Answer 11

ill put up what i have done so far

Answer 12

Someone please catch my mistake :$

Answer 13

Ill show you it step by step I guess

Answer 14

\(\left[\dfrac{f(x)}{g(x)}\right]' = \dfrac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} \)

Answer 15

Someone please catch my mistake :$

Answer 16

\[\frac{ (t^2-1)\prime(t+4)-(t^2-1)(t+4)\prime }{ (t+4)^2 }\]

Answer 17

If you got -1 in the numerator you didnt distribute the negative sign

Answer 18

OOOOH i see your mistake :)

Answer 19

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

Answer 20

and the reason you did not distribute the negative sign is because you have it as a plus sign

Answer 21

Look at what I did

Answer 22

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

Answer 23

Basically you were adding with the quotient rule when it's - :)

Answer 24

Do you see that? you make it f'g+fg'/g2 instead of f'g-fg'/g2

Answer 25

No i dont seee it :(

Answer 26

look at your first step

Answer 27

OH

Answer 28

Omg lol -.-

Answer 29

Haha!

Answer 30

Yes i see it thankssss !!

Answer 31

:)

Answer 32

advice for these kind of questions is

Answer 33

you should always put the rule on the side before you begin differentiating ;)

Answer 34

Yesss I should hahha

Answer 35

\(\left[\dfrac{f(x)}{g(x)}\right]' = \dfrac{g(x)f'(x) {\color{red}{-}} f(x)g'(x)}{[g(x)]^2}\)

Answer 36

thank you @mathstudent55 , i found my mistake :)