Question

I need assistance solving 4 problems. This is a college calculus course..

Answer 1

For this one I got 42/ x^7 but it says its x^8 .. Am I wrong? Or is the system messing up?

Answer 2

Remember that 1/x^6 can be rewritten as x^-6

Does that help?

Answer 3

I know that. Its saying the answer is 8 when im getting 7.

Answer 4

Because d2y/dx2 means the second derivative

x^-6 and that derivative would be with the power rule

-6x^-7

but they want you to take one more derivative: the derivative of -6x^-7

Answer 5

How about I write out what I have and you tell me where I went wrong?

Answer 6

Sure, let's do that

Answer 7

dy/ dx = d/ dx (1/ x^6) = 1 d/dx x^-6

= 1(-6x^-7)
= -6x^-7

d/ dx (-6x^-7)
= 42/ x^7
or 42x^-7

Answer 8

I should have changed it to 8 on the last step right?

Answer 9

Correct
The exponent part of the power rule is always n-1

You're exponent is ^-7, which means according to the power rule the second derivative would be (-7-1) for the exponent

Answer 10

Okay. Can you help with three more?

Answer 11

Yeah, let's see them. Ill see what I can do

Answer 12

I have no clue how to solve this one.

Answer 13

you need to solve the following for x: y'=7

Answer 14

where y=-.025x^2+4x

Answer 15

Huh?

Answer 16

which part are you stuck on?
finding y' given y=-0.025x^2+4x
or solving the equation y'=7?

Answer 17

Um the whole thing. I have no examples for this problem at all. I have one example of it with both x and y points.

Answer 18

I honestly have no idea on where to start.

Answer 19

do you know constant rule, power rule, and addition rule?

Answer 20

for finding derivative?

Answer 21

Uh can you show me what they look like?

Answer 22

we don't really need constant rule for this one unless you want to use product rule to differentiate something in the form (cf(x))'
we do need the following though at the very least
\[(c \cdot f(x))' =c \cdot (f(x))' =c \cdot f'(x) \text{ is constant multiple rule } \\ (x^n)'=n x^{n-1} \text{ is power rule } \\ (f(x)+g(x))'=(f(x))'+(g(x))'=f'(x)+g'(x) \text{ is addition rule }\]

Answer 23

\[y'=(-0.025x^2+4x)' \\ y'=(-0.025x^2)'+(4x)' \text{ by addition rule } \\ y'=-0.025(x^2)'+4(x)' \text{ by constant multiple rule }\]
can you continue to find y'

Answer 24

I kind of recognize power. And addition looks familiar but I dont think Ive used it.

Answer 25

It's kind of like breaking up the equation and doing each derivative serperately

Answer 26

Why do you have dash things behind the parenthesis?

Answer 27

( f(x))' means to take derivative of f(x)

Answer 28

(x^2)' means take derivative of x^2
(x)' means take derivative of x

Answer 29

Um okay.. So x^2 would equal 2x?

Answer 30

k

Answer 31

\[y'=(-0.025x^2+4x)' \\ y'=(-0.025x^2)'+(4x)' \text{ by addition rule } \\ y'=-0.025(x^2)'+4(x)' \text{ by constant multiple rule } \\ y'=-0.025(2x)+4(x)' \text{ by power rule }\]

Answer 32

still need to find (x)'

Answer 33

It will get dropped and stay as 4..?

Answer 34

(x)'=1 so 4(x)'=4(1)=4

Answer 35

\[y'=-0.025(2x)+4(1) \\ y'=-0.050x+4 \]
so solve y'=7 since you are trying to find when slope of tangent line is 7

Answer 36

I got -0.05x.. Is that wrong? You put -0.050x

Answer 37

-.05 is the same as -0.050
or the same as -0.050000000000000000000000000000000000000000000000000000000

Answer 38

Okay.

Answer 39

So set up the equation like:

7=-0.05x +4 ??

Sorry im trying to keep up

Answer 40

yes that's right

Answer 41

Okay so I would minus 4 to the other side. Correct?

Answer 42

minus 4 on both sides

Answer 43

Yeah thats what I meant.. I got -.0166666667 ... Which would change into -0.017? Or -0.02

Answer 44

oh mine for what?

Answer 45

\[7=-0.05x+4 \\ \text{ subtract 4 on both sides } \\ 7-4=-0.05x +4-4 \\ \\ 3=-0.05x\]
you should divide both sides by -0.05

Answer 46

Oh my gosh -.-

Answer 47

x= -60

Answer 48

\[\frac{3}{-0.05}=x \\ -60=x\]
right

Answer 49

now if you want to find corresponding y value you will have to use y=f(x) for that

Answer 50

Next ill plug that in for x in the original equation right?

Answer 51

that is evaluate your original equation at x=-60

Answer 52

y= -330

Answer 53

Awesome

Answer 54

Alright. Thank you. I have 2 more.

Answer 55

I dont know how to solve this one either.

Answer 56

Eww what a messy question and answer xD

This involves a chain rule

Answer 57

I can work a quick example on how to do the chain rule if you'd like? (I got to head to work in a minute)

Answer 58

Okay. That will work

Answer 59

Let's do one like the problem but different:

So you have a polynomial inside of a exponent

(2x^2 + 4x + 6)^3

So what you want to do is pretend that whole polynomial is just x, so it'll be

(x)^3

Answer 60

The derivative of that is quite simple: 3(x)^2

After you take that derivative, replace it with the polynomial to get

3(2x^2 + 4x +6)^2

Answer 61

After this, you're going to want to multiply this thing by the derivative of what's inside the parenthesis

2x^2 + 4x + 6

That derivative being 4x +4

So the final result would be those two things multiplied together as the derivative

(3(2x^2+4x+6)^2) times (4x+4)

Answer 62

That cant be the final step. I got (8(6x^2 -7x+4)^7) (12x-7).. Which isnt the final answer.

Answer 63

Right, because they want the 2nd derivative

Answer 64

Okay. Whats the next step?

Answer 65

[Phone] I would like to help you when I get home because it's confusing. You have to do the product rule while doing chain at the same time

Answer 66

It was mean of them to give this lol

Answer 67

Okay.. Ill post in a new question for now. If no one helps Ill tag you for you to help later. Thank you.

Answer 68

Anytime! Sorry

Answer 69

My professor and Pearson are retriceoles. This is one of the questions from the quiz that I have failed twice now. I have 8 homework assignments done and not one of them are exactly the same to these questions. Youtube has been my friend today.

Answer 70

Really open study you changed a ss- hole to that -.-

Answer 71

Right. Pearson is overpriced bs