Question

help @shadow calc

Answer 1

need 2

Answer 2

@mhchen

Answer 3

isn't it like 180 degrees of the graph sorta

Answer 4

Here's where the derivatives are 0

Answer 5

and there's your slope graph

Answer 6

oh ok

Answer 7

up for 4 more problem? 1 is just an explaining something I have to record myself joy lol

Answer 8

Kayla, Shadow doesn't help with calculus. Try SmokeyBrown or sillybilly123 instead.

Answer 9

@Falconmaster yu said to tag you but I don't know if you do calc

Answer 10

i do geometry but i can try

Answer 11

https://www.mathway.com/Calculushttps://www.mathway.com/Calculushttps://www.mathway.com/Calculushttps://www.mathway.com/Calculus

Answer 12

I have to show my work

Answer 13

it should show you the work -think emoji here-

Answer 14

not for free

Answer 15

hm idk im sorry Brainly?

Answer 16

@mhchen

Answer 17

Okay so problem 5.

Answer 18

Basically find the second derivative of
\[\frac{2}{x}+3\sqrt{x}\]

Answer 19

okay

Answer 20

You can re-write that as:
\[2x^{-1}+3x^{\frac{1}{2}}\]

Are you able to find the first derivative of that?

Answer 21

3/2x^1/2-2/x^2

Answer 22

second derivative is 4/x^3-3/4x^3/2

Answer 23

woah hold on. the first one should be
3/2^(-1/2) - 2/x^2

Answer 24

You see how the exponent is -1/2 instead of 1/2?

Answer 25

you're right

Answer 26

my teacher gave us the answers but basically wants us to show work to get them

Answer 27

so -3/2?

Answer 28

Yeah

Answer 29

but the rest is correct?

Answer 30

4/x^3-3/4x^(-3/2)

Answer 31

okay

Answer 32

then plug in 4?

Answer 33

Yup

Answer 34

I got -1/32

Answer 35

getting there but yes that's the answer lol

Answer 36

I got it lol

Answer 37

onto problem 6. Instantaneous rate of change just means the derivative.

Can you find the derivative of f(x)/x^2

Answer 38

-f/y^2?

Answer 39

it should be like this:
\[\frac{f'(x)x^2 - f(x)(x^2)'}{(x^2)^2}\]

You know the formula for finding the derivative of division stuff right?

Answer 40

oh yeah I do remember that

Answer 41

and then would I plug x in ?

Answer 42

okay so I did something wrong I go 1/3 and need to have 11/27

Answer 43

yah

Answer 44

uh lemme try it myself

Answer 45

I plugged in the appropriate numbers for x f' and f

Answer 46

\[\frac{f'(3)(3)^2-f(3)(2(3))}{(3)^4} = \frac{(5)(9)-(2)(6)}{81}=\frac{45-12}{81}=\frac{33}{81}\]

Answer 47

oh hold on

Answer 48

got it now number 10

Answer 49

one more this one I messed up somewhere

Answer 50

it is supposed to be x-2 not +2

Answer 51

oh uh lemme check 8 first then

Answer 52

I completely forgot how to do 8.

Answer 53

lol

Answer 54

I mean if you plugged in what I have it would still work

Answer 55

let's do 10 lol

Answer 56

This is a secant.

Answer 57

You see how 'a' is the starting point,
and 'h' is how far away it goes

Answer 58

oh I see how 8 is correct

Answer 59

yes

Answer 60

So if you bring 'h' to 0,
basically it becomes like this:

Answer 61

That's a tangent line, AND it's the instantaneous slope at 'a'.

Answer 62

so f(a) is a secant?

Answer 63

just trying to write this down so I can record myself and explain lol

Answer 64

f(a) is just a point.

Look at my first picture.

That line there is a secant, and it's made by
\[\frac{f(a+h)-f(a)}{h}\]

Here is the tangent (and instantaneous slope:

As you can see h is gone because this is
\[lim_{h->0}\frac{f(a+h)-f(a)}{h}\]

Answer 65

is that h->0 in other words say it is going further away from 0?

Answer 66

in the negative

Answer 67

It's going towards 0.

Answer 68

It was positive before. But now it's getting smaller towards 0.

Answer 69

okay was a little confused for a sec but ty for the help!

Answer 70

Here is a beautiful gif: https://commons.wikimedia.org/wiki/File:Tangent_animation.gif

Answer 71

btw Symbolab is great for basic step-by-step walkthroughs of most calculus problems