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Mathematics
kaylak:

calc help @vocaloid

5 months ago
Study2Learn:

Where's the question?

5 months ago
kaylak:

I will post but are you good in calc? I really need help!

5 months ago
Study2Learn:

I'm not good in calc.. however, I am taking the class as of now.

5 months ago
kaylak:

1 attachment
5 months ago
Study2Learn:

What level of calc are you taking?

5 months ago
kaylak:

ab

5 months ago
Study2Learn:

ap calculus?

5 months ago
kaylak:

I would think the graph goes like this|dw:1540222856558:dw|

5 months ago
kaylak:

yes

5 months ago
Study2Learn:

Oh, I'm sorry, I wouldn't be much of help then. I'm taking pre-calc

5 months ago
kaylak:

@kittybasil do you know?

5 months ago
kaylak:

you commented and then deleted it ?

5 months ago
ZoeyBiocth16:

@shadow

5 months ago
kittybasil:

@ZoeyBiocth16 js Shadow doesn't help in calc

5 months ago
ZoeyBiocth16:

ok note to self and thank u kitty

5 months ago
Study2Learn:

Yea, shadow doesn't offer help in that particular course-- you can still ask him if you'd like though.

5 months ago
kaylak:

@mhchen is online! yay!

5 months ago
kaylak:

|dw:1540224158173:dw| is this the derivative?

5 months ago
mhchen:

|dw:1540224157441:dw| It goes up by a constant slope. Then it goes down by a constant slope. Since the slope is the derivative. It's more like this: |dw:1540224191125:dw|

5 months ago
kaylak:

ah okay could you help with 4 more really 2. I have 2 answered

5 months ago
mhchen:

ye

5 months ago
kaylak:

1 attachment
5 months ago
kaylak:

1 attachment
5 months ago
mhchen:

First one is correct.

5 months ago
mhchen:

The second one... \[\sqrt{x} = x^{\frac{1}{2}}\]

5 months ago
kaylak:

2 attachments
5 months ago
mhchen:

Can you find the derivative of \[x^{\frac{1}{2}}\]

5 months ago
kaylak:

1/2x^1/2

5 months ago
kaylak:

1/ 2x ^1/2

5 months ago
mhchen:

\[\frac{1}{2}x^{-\frac{1}{2}}\] Don't forget the negative since 1/2 - 1 = -1/2

5 months ago
kaylak:

so still none of the above?

5 months ago
mhchen:

So the answer for 2 is \[\frac{1}{2x^{\frac{1}{2}}} \] or \[\frac{1}{2 \sqrt{x}}\]

5 months ago
kaylak:

oh okay

5 months ago
mhchen:

For the third question. You can rewrite it as \[8x^{-\frac{1}{2}}\] can you find the derivative of that? Remember you multiply 8 by the exponent (-1/2) and then you subtract 1 from the exponent.

5 months ago
kaylak:

-4x^-3/2

5 months ago
kaylak:

right?

5 months ago
mhchen:

Yup. Now plug x=4 in for x to find f'(4)

5 months ago
kaylak:

-1/2 or d

5 months ago
mhchen:

Yeah

5 months ago
kaylak:

last one!

5 months ago
kaylak:

find the derivative?

5 months ago
mhchen:

Okay do you remember this formula: \[(\frac{a}{b})' = \frac{a'b - ab'}{b^2}\]

5 months ago
kaylak:

yes

5 months ago
mhchen:

So yeah same thing: \[\frac{x^2}{f(x)} \] can you try and find the derivative of that

5 months ago
kaylak:

1/Y?

5 months ago
kaylak:

sorry caps lock lol

5 months ago
mhchen:

er i'm not sure where you got Y from.. \[(\frac{x^2}{f(x)})' = \frac{(x^2)'f(x) - (x^2)(f'(x))}{f(x)^2}\] it should be like this instead. Do you see what I did?

5 months ago
kaylak:

oh yeah I needed to put in that equation

5 months ago
kaylak:

makes sense since I have to plug in values

5 months ago
mhchen:

Yeah, just plug in the values for f(x) and f'(x) in that thing

5 months ago
kaylak:

isn't it false

5 months ago
kaylak:

unless I did something wrong I'm checking now

5 months ago
mhchen:

uh let me check.. 6*2-9*4 / 4 = 12 - 36 / 4 = -24/4 = -6

5 months ago
mhchen:

yeah I think it's false.

5 months ago
kaylak:

hey that's what I got

5 months ago
mhchen:

welp that's it then

5 months ago
kaylak:

yay ty!

5 months ago
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