Question

3x^3+2x^2 +X+5 the derivative
evalauate y' when x=3 need a little help bit rusty

Answer 1

in order to the first you lower the power by one like this

3x^2+2x^1+5 that is der right?

Answer 2

you also have to divide by the original index

Answer 3

you divide by 3 because of 3x^3

Answer 4

x^n=nx^n-1

Answer 5

so if I divide by 3, than that would make this x^2 instead of 3x^2 I am on the right track?

Answer 6

opps ive got mixed up

Answer 7

y' 3x^2 is X^2 right?

Answer 8

i should have said times by the index

\[y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=anx^{n-1}\]

Answer 9

so sorry if i confused you

Answer 10

so times by index makes 3x^3 Y' 6x^2 right?

Answer 11

not quite
\[y(x)=3x^3\qquad y'(x)=3\times3x^2\]

Answer 12

here is what I know y' when doing this you lower the power by one and something??

Answer 13

when taking the derivative multiply by the power and then lower the power

Answer 14

so you are telling that power before lowering divide to y'

Answer 15

like this 4x^3 so that would be 'y 4 right?

Answer 16

4x

Answer 17

i ment times when i said divide originally , dont divide when taking the derivative

Answer 18

can we start over plz..

Answer 19

yes ,

Answer 20

y' 4x^2 would y'4x

Answer 21

\[y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=n\times ax^{n-1}\]
\[y(x)=4x^2\qquad\Rightarrow\qquad y'(x)=2\times 4x^{2-1}\]

Answer 22

y'(x) would be 8x

Answer 23

thats right now

Answer 24

okay so the power is what you are times before lowering the power?

Answer 25

that is right

Answer 26

so for another example 10X^5 y'(x) 50x^4

Answer 27

you got it \(\checkmark\)

Answer 28

so back to the problem now

Answer 29

\[y(x)=3x^3+2x^2 +x+5\]\[y'(x)=\]

Answer 30

y'(x) 6x+4X +5

Answer 31

not quite,
note that
\[x=x^1\]\[5=5x^0\]

Answer 32

y'(x) 9x+4+5

Answer 33

try again

Answer 34

y'(x) 6x^2 +4x+5 +1

Answer 35

ok lets do this one term at a time

\[(3x^3)^\prime=\]

Answer 36

y'(x) 9x^2

Answer 37

right,
now
\[(2x^2)^\prime=\]

Answer 38

y'(x) 9x^2 +2x

Answer 39

y'(x) 9x^2 4+5+1

Answer 40

nope

Answer 41

y'(x) 9x^2 +4X+5

Answer 42

you have the first two terms right,

Answer 43

\[f(x)=x=x^1\]
\[f^\prime(x)=1\times x^{1-1}=x^0=1\]

Answer 44

so y'(x) 9x^2+4x+1+5

Answer 45

the derivative of the last term
\[g(x)=5=5x^0\]
\[g^\prime(x)=0\times 5x^{0-1}=\]

Answer 46

is 0

Answer 47

yeah the derivative of a constant is always zero

Answer 48

so the end number unless it has a power drop off?

Answer 49

thats right

Answer 50

\[f(x)=x^2+5\qquad f'(x)=2x\]\[g(x)=x^2+9\qquad g'(x)=2x\]\[h(x)=x^2−71\qquad f'(x)=2x\]

Answer 51

that last one is ment to be h'

Answer 52

cool love it so far because in math you can't drop anything, in calc 1 you can this is way cool..

Answer 53

do you have time to do one more?

Answer 54

yeah can you get the answer to you original question now?

Answer 55

\[y(x)=3x^3+2x^2 +x+5\]
\[y~'(x)=\]

Answer 56

y'(x) 9x^2+4x+1

Answer 57

\[\Large\color\red\checkmark\]

Answer 58

okay any other tips here?

Answer 59

\[y'(x)=\frac{\text dy(x)}{\text dx}\]

Answer 60

explain a little??

Answer 61

its just a different notation, they men the same thing

Answer 62

the second has a bit more information because it explicitly say that we are to take the derivative of the function with respect to x, ie we look at the power of the x terms,

Answer 63

dx(x)^1?dx

Answer 64

so the power on the is 1 right?

Answer 65

just to let know, I am in college not high school

Answer 66

\[y(x)=x=x^1\]
\[\frac{\text dy(x)}{\text dx}=y'(x)=1\times x^{1-1}=x^0=1\]

Answer 67

im not sure what you mean by this
dx(x)^1?dx

Answer 68

I meant to use / not?

Answer 69

dx(x)^1/dx \[=\large \frac{\text d(x)^1}{\text dx}\]?

Answer 70

yes that is what I meant to type

Answer 71

\[\frac{\text d(x)}{\text dx}=\frac{\text d(x^1)}{\text dx}=1\times x^0=1\]

Answer 72

im not sure if i am answer your question?

Answer 73

okay now for the last question here how do this with y"(x) sin

Answer 74

yes you have

Answer 75

should we evaluate y' in the first question when x=3 before that?

Answer 76

y'(x) 9x^2+4x+1 so plug in 3 so we have y'(x) 9x^2 (3) +4x(3) +1 which is y'(x) 27x^2+ 12X +1

Answer 77

\[y(x)=3x^3+2x^2+x+5\]
\[y'(x)=9x^2+4x+1\]

\[y'(3)=9(3)^2+4(3)+1=9\times3\times3+4\times3+1\]

Answer 78

y'(3) 81x^2+12+1

Answer 79

12x

Answer 80

when you evaluate a function at a particular x value you substitute this value for x in the equation, there should be no x in left when evaluated

Answer 81

sorry i mess me up by not putting the like this (3)^2

Answer 82

so the answer is F'(3) 81x^2 + 12X +1