Question

Find the derivative of the function using the definition of derivative.

Answer 1

bunch of algebra no doubt

Answer 2

\[g(t) = \frac{ 3 }{ \sqrt{t}}\]

Answer 3

Just a couple a I'm having trouble with.

Answer 4

yup bunch of algebra
\[\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\]

Answer 5

so first of \(g(t+h)=\frac{1}{\sqrt{t+h}}\) right?

Answer 6

yes

Answer 7

k lets work only in the numerator, forget bout the \(h\) in the denominator for the moment, we can put that in last

Answer 8

\[\frac{1}{\sqrt{t+h}}-\frac{1}{\sqrt t}\] is what we need to work with i.e. actually perform the subtraction

Answer 9

\[\frac{ 3 }{ t+h } -(3/\sqrt{t})\]

Answer 10

where did the 3 come from ?

Answer 11

oh nvm it is there, i forgot about it
we could leave that out too, but lets put it in

Answer 12

\[\frac{3}{\sqrt{t+h}}-\frac{3}{\sqrt t}\] is the first step
then subtract

Answer 13

you good with that, or need help to subtract?

Answer 14

\[3\sqrt{t}-3\sqrt{t+h}/(\sqrt{t+h)}(\sqrt{t)}\]

Answer 15

right

Answer 16

now the gimmick is to multiply top and bottom by the conjugate of the numerator, i.e. rationalize the numerator

Answer 17

OK

Answer 18

the 3 is annoying, lets factor it out of the whole damn thing and compute
\[3\left(\frac{\sqrt{t}-\sqrt{t+h}}{\sqrt{t}\sqrt{t+h}}\right)\times \left(\frac{\sqrt{t}+\sqrt{t+h}}{\sqrt{t}+\sqrt{t+h}}\right)\]

Answer 19

Ok

Answer 20

leave the denominator in factored from, and after doing the algebra note that the numerator is only \(-h\)

Answer 21

So we have -h as the numerator and a lot of stuff for the denominator g(t) =
3
t

Answer 22

right
now divide by \(h\) (the original \(h\) that we ignored in the definition) and you are left with
\[\frac{-3}{\sqrt{t}\sqrt{t+h}(\sqrt{t}+\sqrt{t+h})}\]

Answer 23

finally to take the limit, replace each \(h\) by \(0\) and you are done

Answer 24

\[t+2\sqrt{t}\sqrt{t+h}+t+h\]

Answer 25

no \(h\) should survive
they are all zero

Answer 26

\[\frac{-3}{\sqrt{t}\sqrt{t+h}(\sqrt{t}+\sqrt{t+h})}\] make it \[\frac{-3}{\sqrt{t}\sqrt{t}(\sqrt{t}+\sqrt{t})}\]

Answer 27

Ok

Answer 28

The answer given is \[\frac{ -3 }{ 2t ^{\frac{ 3 }{ 2 }} }\]

Answer 29

yes that is what you get when you compute the thing above

Answer 30

Great..I will review. appreciate the step by step instructions.

Answer 31

\[\sqrt{t}\sqrt{t}=t\]
\[\sqrt{t}+\sqrt{t}=2\sqrt{t}\] you get
\[2t\sqrt{t}=2t^{\frac{3}{2}}\] for the denominator

Answer 32

yw