calc help @vocaloid

5 months agoWhere's the question?

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

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

5 months agoWhat level of calc are you taking?

5 months agoab

5 months agoap calculus?

5 months agoI would think the graph goes like this|dw:1540222856558:dw|

5 months agoyes

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

5 months ago@kittybasil do you know?

5 months agoyou commented and then deleted it ?

5 months ago@shadow

5 months ago@ZoeyBiocth16 js Shadow doesn't help in calc

5 months agook note to self and thank u kitty

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

5 months ago@mhchen is online! yay!

5 months ago|dw:1540224158173:dw| is this the derivative?

5 months ago|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 agoah okay could you help with 4 more really 2. I have 2 answered

5 months agoye

5 months agoFirst one is correct.

5 months agoThe second one... \[\sqrt{x} = x^{\frac{1}{2}}\]

5 months agoCan you find the derivative of \[x^{\frac{1}{2}}\]

5 months ago1/2x^1/2

5 months ago1/ 2x ^1/2

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

5 months agoso still none of the above?

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

5 months agooh okay

5 months agoFor 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-4x^-3/2

5 months agoright?

5 months agoYup. Now plug x=4 in for x to find f'(4)

5 months ago-1/2 or d

5 months agoYeah

5 months agolast one!

5 months agofind the derivative?

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

5 months agoyes

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

5 months ago1/Y?

5 months agosorry caps lock lol

5 months agoer 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 agooh yeah I needed to put in that equation

5 months agomakes sense since I have to plug in values

5 months agoYeah, just plug in the values for f(x) and f'(x) in that thing

5 months agoisn't it false

5 months agounless I did something wrong I'm checking now

5 months agouh let me check.. 6*2-9*4 / 4 = 12 - 36 / 4 = -24/4 = -6

5 months agoyeah I think it's false.

5 months agohey that's what I got

5 months agowelp that's it then

5 months agoyay ty!

5 months ago