Question

Need help with derivatives and the definition of derivatives..

Answer 1

The ones with a red line through them, I've already solved. Thank you in advance!

Answer 2

are you using the f(x+h) - f(x) divided by (h) method?

Answer 3

Yes! Sorry for the delay, my internet is not doing well

Answer 4

well let's set it up

Answer 5

do you know how to get rid of the radicals?

Answer 6

(going to bed soon but here's the solution http://www.sosmath.com/calculus/diff/der01/der01a.html)

tag someone if you need an explanation ^^

Answer 7

\[\lim_{h \rightarrow 0} \frac{ f(x+h)-f(x) }{ x+h }\]

Answer 8

@LifeIsADangerousGame

say we start of like this. right?

\[slope = \frac{ y }{ x }\]

Answer 9

we have some function x and we increase our x value by a small number h

so we would have x and x+h

now

the y value would be f(x) and f(x+h)

think of it as taking the difference between these two values and finding the limit as h gets really small or approaches zero.

Answer 10

I started the problem but it wasn't working out...(trying to type what I've done so far in here but it's taking a while, but I understand what you're doing so far)

Answer 11

\[f(x)= 3\sqrt{x}\] \[f \prime = \frac{3\sqrt{x + h} - 3\sqrt{x}}{ h }\]

Answer 12

\[\frac{ 3\sqrt{x+h} - 3\sqrt{x} }{ 3\sqrt{x}+h }\]

Answer 13

we could probably rationalize the numerator then simplify

Answer 14

how did the 3sqrt(x) get on the denom as well?

Answer 15

it should just be h at the bottom

Answer 16

was typing fast

Answer 17

Okay, so rationalizing the numerator, I'll do that now

Answer 18

9h/h(3sqrt(x + h) + 3sqrt(x)?

Answer 19

i see now

Answer 20

\[\frac{ 3\sqrt{x+h}-3\sqrt{x} }{ h} * \frac{ 3\sqrt{x+h} + 3\sqrt{x} }{ 3\sqrt{x+h}+3\sqrt{x} }\]

Answer 21

Right, that's what I meant sorry

Answer 22

\[\frac{ 9(x+h)-9(x) }{ h(3\sqrt{x+h} +3\sqrt{x} )} \]

Answer 23

\[\frac{ \cancel\9x+9h-\cancel\9x }{ h(3\sqrt{x+h} + 3\sqrt{x} } = \frac{ 9\cancel\h }{\cancel\h(3\sqrt{x+h}+3\sqrt{x} }\]

Answer 24

now watch what happens when we plug in h = 0

Answer 25

\[\frac{ 9 }{ 3\sqrt{x}+3\sqrt{x} } = \frac{ 9 }{ 6\sqrt{x} } = \frac{ 3 }{ 2\sqrt{x}}\]

Answer 26

so we can actually re-write this too

Answer 27

\[\frac{ 3 }{ 2*\sqrt{x} }*\frac{ 2\sqrt{x} }{2\sqrt{x} } = \frac{ 6\sqrt{x} }{ 4x } = \frac{ 3\sqrt{x} }{ 2 x}\]

Answer 28

I could have stopped before but if you don't think this is the same we can use power rule

\[\frac{ 3 }{ 2 }\frac{ x^{0.5} }{ x^{1} } = x^{0.5-1} = \frac{ 3 }{ 2 } x^{-0.5} ~or~ \frac{ 3 }{ 2\sqrt{x} }\]

different flavors of the same answer.

Answer 29

no that makes sense! Thanks for going through all of it so in depth. I finally understand it

Answer 30

yeah, you'll learn the power rule of differentiation which makes this much easier.

Answer 31

Yeah, he taught us that already, but the exam is on the definition so I have to get used to using both