Question

Q 11, 12 ! Not sure if Im doing it right,

Answer 1

Your first answer of -5 is indeed correct!

Answer 2

omg yes !

Answer 3

\[\large f(x) = -5x - 8\]

The derivative of a term with the variable is just the term, and the derivative of a constant is 0...so yes

\[\large f'(x) = -5\]

Answer 4

is b 12?

Answer 5

Now quite...for that we will use the power rule

\[\large f(x) = 2x^2 + 4x\]

The power rule says that

\[\large f(x) = ax^n\]
\[\large f'(x) = nax^{n - 1}\]

Lets focus on just the 2x^2 for a minute

\[\large f(x) = 2x^2\]
n = 2
a = 2
so
\[\large f'(x) = 2\times 2 x^{2 - 1}\]
\[\large f'(x) = 4x\]

Answer 6

So that is something good to memorize

The power rule, you multiply the power...times the coefficient of the 'x'.....then subtract 1 from the power

For example:
\[\large f(x) = 4x^6\]
\[\large f'(x) = 6\times 4x^{6 - 1}\]
\[\large f'(x) = 24x^5\]

Answer 7

oh okay, when do we use this rule and the other rules? How do we know

Answer 8

So using that...along with the fact that the derivative of a term with the variable is just the term.....we have

\[\large f(x) = 2x^2 + 4x\]

We just found the derivative of 2x^2 to be 4x....and the derivative of the + 4x will be just 4 right? so we have

\[\large f'(x) = 4x + 4\]

Answer 9

ohh alright ! and the last 4 is just 4 because 4x, the x itself is equal to one?

Answer 10

And you use this rule whenever you see a term, with a variable, raised to and exponent

Just like the example I gave you

\[\large f(x) = 4x^6\]

Compare that to the formula
\[\large f(x) = ax^n\]
means that a = 4...n = 6

so

\[\large f'(x) = n\times ax^{n-1}\]
\[\large f'(x) = 6 \times 4x^{6 - 1}\]
\[\large f'(x) = 24x^5\]

Answer 11

and yes, the 4x is just 4 because the derivative of x is = 1

Answer 12

Okay got it ! would we just leave it at 4x+4 then ?

Answer 13

Correct! That is the derivative :)

Answer 14

Wow thanks !! :)

Answer 15

Anytime! :)

But on to Q12

What is a?

Answer 16

I got 4

Answer 17

for b) I got 3x^2-2x
c) -2x+5 d) 1 1/3

Answer 18

A,B and C PERFECT!

D)

\[\large \sqrt[3]{x}\]

Remember that is written as

\[\large x^{1/3}\]

Looks like a power rule again right? Just with the coefficient of 'x' being 1...

Answer 19

e) y=1+18 sqroot x^2 f) undefined

Answer 20

oh yes okay !

Answer 21

Alright, so lets back up to D here...what do we get this time?

Answer 22

x^ 1/3 , we dont turn the x into a 1

Answer 23

Hmm not quite, we'll break this one down

\[\large x^{1/3}\]

again, we can compare it to the

\[\large f'(x) = n\times ax^{n-1}\]

n here will be 1/3 ...and 'a' will just be 1 (since 'x' has a coefficient of 1)

so

\[\large f'(x) = 1/3 \times x^{1 - 1/3}\]
\[\large (1/3)x^{2/3}\]

also can be written as

\[\large \frac{x^{2/3}}{3}\]

Answer 24

wouldn't it be -2/3?

Answer 25

isn't it (1/3) x x^1/3-1

Answer 26

Yay you caught my mistake :D I was waiting for so long to see if you would :DDDD

Answer 27

Thousand medals if I could :D lol

Answer 28

Yay wow thanks ! :D

Answer 29

were e and f okay ?

Answer 30

lol no problem :)

alright so did that make sense?

And well lets check them..did you use this same method for those?

Answer 31

I believe so. For e I got 1+18 sqroot x^2, and f) undefined ...

Answer 32

oh wait I think 2 may be wrong

Answer 33

and f

Answer 34

ill try again

Answer 35

Let me know if you need help :)

Answer 36

would e be y= 1/2 + 9x^1/2 ?

Answer 37

in the starting I got y=x^1/2+6x^3/2 then went on from there

Answer 38

and I still got undefined for f

Answer 39

Well lets see...how did x^1/2 become just 1/2?

Remember the power rule...

it isn't 1/2x ....if that were the case then yes it would be 1/2...

but this is instead x^1/2

Answer 40

no wait

Answer 41

Well lets focus on e) for now :)

Answer 42

oh okay wait I made a mistake, I ll try one more time and show you my work !

Answer 43

So close, would be perfect but just 1 mistake...can you spot it?

*hint, exponents*

Answer 44

Ummm the second line where its suppose to be 6.x^1/3?