Question

I have the answer can someone check to see if I;m right
Which of the following is f ' (x) for f(x)=-3x^-5

Answer 1

What's your answer?

Answer 2

-15x^-4

Answer 3

Sorry, that's incorrect.

Answer 4

\[\frac{d}{dx}[u^n] = nu^{n-1}\]

Answer 5

Ok so the is the answer 12x^-4

Answer 6

No. What does n = ?

Answer 7

n=-3

Answer 8

No, isn't it \[f(x) = -3x^{-5}\]?

Answer 9

yes so then doesn=-5

Answer 10

Yes!

Answer 11

Don't worry about the -3 in front for the moment, what is the derivative of \(x^{-5}\)?

Answer 12

-4

Answer 13

No...it's going to be some function of x

Answer 14

Like-6

Answer 15

No...it's not just a number, it needs to have an x in there!

Answer 16

oh ok I gotha so like -8x

Answer 17

No....

\[\frac{d}{dx}[x^n] = nx^{n-1}\]Your value of n is n = -5, right?
\[\frac{d}{dx}[x^{-5} = -5x^{-5-1} = 5x^{-6}\]

Answer 18

ok so then what or is the 5x^-6 my answer

Answer 19

Nuts, missed a bracket:
\[\frac{d}{dx}[x^{-5}] = -5x^{-5-1} = -5x^{-6}\]

So what does \[\frac{d}{dx}[-3x^{-5}] = \]?

Answer 20

ok so I put it into my calculator bt for the space in the calculator where it says x= at the end what do I put

Answer 21

You don't need a calculator for this...

If the derivative of x^-5 = -5x^-6, what is the derivative of -3x^-5?

Answer 22

This is where I get confused can you explain this to me

Answer 23

Remember, the definition of the derivative is

\[\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\]

Let's say our f(x) is simply f(x) = x. Then by the definition, we have:

\[\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} =\lim_{h\rightarrow0}\frac{(x+h)-(x)}{h} = 1 \]
right?

Answer 24

yes

Answer 25

Now let's say our f(x) is f(x) = 3x.

\[\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} =\lim_{h\rightarrow0}\frac{3(x+h)-3(x)}{h} =\]\[3 \lim_{h\rightarrow0}\frac{(x+h)-(x)}{h} = 3*1 = 3 \]
right?

Answer 26

so then would the answer be -15x^-6

Answer 27

what is -3 * -5?

Answer 28

watch out forehead, here comes palm :-)

Answer 29

positive 15x^-6

Answer 30

yes!

Answer 31

ok thanks

Answer 32

So I wasn't sure if your initial mistake was subtracting 1 from the leading -3 instead of the exponent, or if you were adding 1 instead of subtracting one from the -5...

Answer 33

I think it was where you had o subtract 1 from -3

Answer 34

Here's another one:

if f(x) = 3x^2, what is f'(x) =?

Answer 35

would it be 6x

Answer 36

correct again! One more:

f(x) = -1x^-1, f'(x) = ?

Answer 37

1x

Answer 38

Nope. Try again.

Answer 39

x^-2

Answer 40

Yes. Remember, the exponent always decreases.

Answer 41

ok but I have another problem but with a fraction can you help me with that one

Answer 42

Sure, as long as it is a quick one...

Answer 43

yes ok same instructions but f '(x) for f(x)=(1/x^7)

Answer 44

\[f(x) = \frac{1}{x^7}\]?

Answer 45

yes

Answer 46

is the answerx^-7

Answer 47

Remember the property of exponents:
\[a^{-n} = \frac{1}{a^n}\]

Answer 48

ok so then the answer would be x^-7

Answer 49

How can you rewrite your problem given that?

Answer 50

what I;m confused

Answer 51

\[f(x) = \frac{1}{x^7} = x^{-7}\]

Answer 52

Now take the derivative

Answer 53

-7x^6

Answer 54

Okay, you're not following the rules:

\[\frac{d}{dx}[u^n] = nu^{n-1}\]

Answer 55

(1/-7x^6

Answer 56

I should be writing that with a du, not a dx, but whatever...

If your initial exponent is n = -7, then the result is \(-7x^{-7-1} = -7x^{-8}\)

Answer 57

oh ok for some reason I was doing -7+1 and that was where I was getting the -6 from

Answer 58

Yes, I figured as much. So, don't do that :-)

If your answer is \(-7x^{-8}\) what would that be written as a fraction?

Answer 59

(-7/x^8)

Answer 60

is that correct

Answer 61

Yes. I was hoping you wouldn't make the mistake of writing 1/(-7x^8) :-)

Answer 62

no

Answer 63

thanks

Answer 64

You're welcome. Remember, watch the subtraction from the exponent carefully!

Answer 65

ok thnks