Question

find the derivative of (4/x) using the defintion of derivative

Answer 1

The definition of derivative (one of them) is\[f'(x)=\lim_{h\to0} \frac{f(x+h)-f(x)}{h}.\] Plugging into this, we need to calculate \[\lim_{h\to0} \frac{4/(x+h)-(4/x)}{h}\]

Answer 2

do i take out both 4/x on the tops or do i have to foil first .. i dont know how to foil a fraction

Answer 3

Sorry, made a little mistake in my work. But you'll want to add the two fractions in the top.

Answer 4

4x^2+h?

Answer 5

\[\frac{4}{x+h}-\frac{4}{x}=\frac{4x-4(x+h)}{x(x+h)}=\frac{4h}{x(x+h)}\]

Answer 6

Then, if we divide the whole thing by h, we get\[\large \lim_{h\to0}\frac{\frac{4h}{x(x+h)}}{h}=\lim_{h\to0}\frac{4h}{hx(x+h)}=\lim_{t\to0}\frac{4}{x(x+h)}\]

Answer 7

Sorry, with a minus sign in front of that since \[\frac{4}{x+h}-\frac{4}{x}=-\frac{4h}{x(x+h)}\]

Answer 8

would you say that is the final answer ? -4 over x(x+h)

Answer 9

why i am asking if because my teacher wants us to do it the long way (definiton of derivitve) and the short cut way... but we got two different answers? because i did 4/x the short cut way and got -4/x^2

Answer 10

Not yet. We need to take the limit of that as h goes to 0. But since h=0 does not give you an undefined fraction, you can just plug in h=0 to the final limit.

Answer 11

So\[\lim_{h\to0}-\frac{4}{x(x+h)}=-\frac{4}{x(x+0)}=-\frac{4}{x^2}\]

Answer 12

ok so we did get the sane answer.... yes. thank you so much. i will study this.

Answer 13

also.. can i ask you one more question? limit with a radical on top.?

Answer 14

Sure.

Answer 15

the teacher told us he is going to give us a problem on the test like that. like limit approaching 3 and one with a radical on the numerator. so can you maybe quiz me on one and then help me solve?

Answer 16

Perhaps something like finding the derivative of \(3\sqrt{x}\). I'm not sure if this will be easy or not, but we can try.

Answer 17

no... something like if the limit = 3 and he gives us x square root of 21 over 7

Answer 18

So just something like\[\lim_{x\to3}\frac{x\sqrt{21}}{\sqrt{7}}?\]

Answer 19

yes

Answer 20

but the denominator (7) isnt in a square root

Answer 21

Alright, so\[\lim_{x\to3}\frac{x\sqrt{21}}{7}.\]This is actually very straightforward. If you just plug in \(x=3\), you get \[\frac{3\sqrt{21}}{7}.\] Since this is perfectly well defined, this is the entire limit.

Answer 22

can it be simplified anymore?

Answer 23

would i multiply 21 by 3?

Answer 24

Since the 3 is not under the square root, you can't just multiply those together. In fact, this is probably the most simplified form you'll get.

Answer 25

so it would be 63/7 or would you say leave it like it was

Answer 26

Leave it like it was.

Answer 27

so what if it was like limit approaching 3 numerator was square root of x+21 denominator 7

Answer 28

\[\lim_{x\to3}\frac{\sqrt{x+21}}{7}\]Again, we first just plug in x=3 and see what we get.\[\frac{\sqrt{3+21}}{7}=\frac{\sqrt{24}}{7}=\frac{2\sqrt6}{7}\]Since this well defined, this is the final limit.

Answer 29

awesome thank you. !!

Answer 30

You're welcome.