Question

Show how to find the derivative of f(t)= (t^2+1)/t using the quotient rule. Then show how to find the derivative of the same function without using the quotient rule and first changing the form of the function. I get two different answers when I do this! HELP MEEEEEEEE

Answer 1

1/t = t^-1

Answer 2

\[f'(t)=\frac{(t^2+1)'t-(t^2+1)(t)'}{t^2}\]

Answer 3

\[[t+t^{-1}]'=1-t^{-2}\]

Answer 4

\[=\frac{(2t+0)t-(t^2+1)(1)}{t^2}\]

Answer 5

\[=\frac{2t^2-(t^2+1)}{t^2}=\frac{2t^2-t^2-1}{t^2}=\frac{t^2-1}{t^2}=\frac{t^2}{t^2}-\frac{1}{t^2}\]
same as what amistre has

Answer 6

\[[t+t^{-1}]'=1-t^{-2}\]
\[[t+t^{-1}]'=\frac{t^2-1}{t^{2}}\]

Answer 7

1-t^{-2}

Answer 8

or he can rewrite his to match mine

Answer 9

lol yay!!

Answer 10

this a repeat?

Answer 11

its all a repeat ;)

Answer 12

No, it's not a repeat, on the first one, I didn't put the numerator in parenthesis, so it was done wrong.

Answer 13

no they are talking about the question

Answer 14

they seen it before
and i think i seen it too

Answer 15

I don't understand how to do it without the quotient rule though! Can someone explain that!

Answer 16

@nls11, it is a good thing to remember (memorize if you will) three fact:
\[\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}\]
\[\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\]and their chain rule cousins
\[\frac{d}{dx}\frac{1}{f(x)}=-\frac{f'(x)}{f^2(x)}\] and
\[\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\]

Answer 17

nls, that what amistre was doing above he wrote as two fractions

Answer 18

math teachers love putting these on exams, and it saves you from rewriting everything using stupid exponents and then using the power rule and getting rid of the exponents

Answer 19

\[\frac{t^2+1}{t}=\frac{t^2}{t}+\frac{1}{t}\]

Answer 20

\[=t+\frac{1}{t}\]
then we find the derivative of that since this is what your question asks