Question

Can somone please help me with this derivative problem. please.
Directions:
use the figure below to estimate each derivative.
If derivative does not exist type DNE

Answer 1

\[g'(2)\] does not exist because of the corner there

Answer 2

ok

Answer 3

\[h'(x)=f'(g(x))\times g'(x)\]

Answer 4

so
\[h'(3)=f'(g(3))\times g'(3)\] do you know what \(g'(3)\) is ?

Answer 5

6

Answer 6

no

Answer 7

I am really confused can you explain it too me please

Answer 8

For g'(2), the derivative doesn't exist because it's a corner necessarily. It doesn't exist because the derivative (in simplest terms the tangent line at the point) is actually undefined (probably something divided by zero). This is an absolute value function, and the derivative does not exist at the turning point.

Answer 9

\(g'(3)\) is the slope of the line
it looks to me like the slope of the line \(g\) at 3 is 2

Answer 10

oh ok so when you said g'(3) look at the graph and then go up and then it equals 2

Answer 11

ok lets go slow

Answer 12

ok

Answer 13

you need \(g'(3)\)

Answer 14

right

Answer 15

and the graph of \(g\) itself is a line

Answer 16

Remember, g'(3) just means the rate of change instantaneously at that point. You can calculate that by looking at the tangent line, which is literally that part of the graph (linear). Use slope formula and calculate the derivative there.

Answer 17

the slope of a line is a constant.
in this case you compute the slope of the line by your eyeballs

Answer 18

ok

Answer 19

you can see that for the graph of the line \(g\) the slope is 2, because the line travels up 2 units for every one unit in the \(x\) direction

Answer 20

ok

Answer 21

that means \(g'(3)=2\)

Answer 22

oh ok cause it goes up 2

Answer 23

over one, up two
right

Answer 24

now we need \(g(3)\) which you read directly from the graph

Answer 25

okk

Answer 26

you can see that when \(x=3\) we have \(g(x)=2\)

Answer 27

it is just a coincidence that they are the same number

Answer 28

right cause you have to go up 2

Answer 29

you see the point \((3,2)\) on the graph of \(g\)

Answer 30

yes

Answer 31

now lets look again at \[h'(3)=f'(g(3))\times g'(3)\]

Answer 32

we know \(g(3)=2\) and also \(g'(3)=2\) so we have
\[h'(3)=f'(2)\times 2\]

Answer 33

so then h'(3)=4

Answer 34

no not yet

Answer 35

oh ok

Answer 36

we have to find \(f'(2)\)

Answer 37

so far we are only here
\[h'(3)=f'(2)\times 2\]

Answer 38

\(f'(2)\) is the slope of the line of \(f\)

Answer 39

which you again compute by your eyeballs
the line goes to the right 4 and down 1, so that slope is \(-\frac{1}{4}\)

Answer 40

okI see that

Answer 41

now we are ready to compute

Answer 42

\[h'(3)=f'(g(3))\times g'(3)=f'(2)\times 2=-\frac{1}{4}\times 2\]

Answer 43

so then it equals -0.5

Answer 44

yes it does

Answer 45

so thenh'(3)=-0.5?

Answer 46

yes

Answer 47

How do you figure out the m'(1)

Answer 48

you need the product rule

Answer 49

oh ok think I got it thanks

Answer 50

\[m(x)=xg(x)\]
\[m'(x)=g(x)+xg'(x)\] so
\[m'(1)=g(1)+1\times g'(1)\]

Answer 51

you need to find both \(g(1)\) and \(g'(1)\)

Answer 52

g(1)=1 and then g'(1)=2

Answer 53

no i don't think so

Answer 54

what is \(g(1)\)?

Answer 55

you have to find this from the graph
look at the x axis, find 1, then go up to the graph of \(g\) and find the corresponding \(y\) value

Answer 56

it looks to me like \(g(1)=2\)

Answer 57

ok so then the g'(1)= 2.5

Answer 58

don't forget that
\(g'(1)\) is the slope of the line of \(g\)

Answer 59

oh right

Answer 60

that slope is pretty clearly negative, because the graph of \(g\) is headed down on that side

Answer 61

ok so like -1

Answer 62

no i don't think it is minus one
for each one step in the x direction (right) how many steps does the function go down?

Answer 63

2

Answer 64

right
so the slope is?

Answer 65

2

Answer 66

gotta run, good luck
but i hope it is clear that the slope is \(-2\) because the line is going down, not up

Answer 67

ok thanks