Question

find the derivative:
y =(2x-7)^3
and y= (3x^2+1)^4

Answer 1

use power rule along with chain rule

Answer 2

hmm so for the first example, we do this?

Answer 3

\[y =(2x -7)^{3}\]
\[\frac{ dy }{ dx }=3(2x -7)^{2}(2)\]
\[\frac{ dy }{ dx }=6(2x -7)^{2}\]

Answer 4

what was that formula? i got up to 3(2x-7)^2

Answer 5

Use chain rule, you still need to derive what is in the parentheses.

Answer 6

\[y =(3x ^{2}+1)^{4}\]
\[\frac{ dy }{ dx }=4(3x ^{2}+1)^{3}(6x)\]
\[\frac{ dy }{ dx }=24x(3x ^{2}+1)^{3}\]

Answer 7

Did you understand?

Answer 8

dy/dx=3(2x−7)^2 (2)

Answer 9

i look at that and i understand where the 3 comes from, but not the 2 on the right

Answer 10

The rule you're using is called the power of a function rule (a form of the chain rule). If\[y =[f(x)]^{n}\]then\[\frac{ dy }{ dx }=n[f(x)]^{(n -1)}[f \prime(x)]\]In other words, multiply the power (exponent) of the function by the number in the front, reduce the power of the function by one, and multiply by the derivative of the function.

In other words, multiply the exponent that the bracket is raised to by the m=number in front of the bracket, then reduce the exponent of the bracket by one, then multiply by the derivative of the function inside the bracket.

Kapish?!

Answer 11

m=number is just a typo so ignore that. it's just 'number'. OK?

Answer 12

So do you understand? Comprendre?

Answer 13

so for example, 3(9x-4)^4 would be 4(3)(9x-4)^3 (9)?

Answer 14

Excellent!

Answer 15

yay :D

Answer 16

so i take the 4, multiply it with the coefficient outside of the parenthesis, subtract 1 from the exponent, find the derivative of 9x-4 which is just 9. gotcha.

Answer 17

Beautiful! Just simplify by multiplying all the monomials so for example your simplified answer should be\[\frac{ dy }{ dx }=108(9x -4)^{3}\] in your example. Alright?

Answer 18

so for 2(x^2-1)^3 i would get 3*2(x^2-1)^2(1)

Answer 19

yep, sure did. i was just trying to nail the basic part of it

Answer 20

Well you could do this without the chain rule if you haven't learnt it but it would take a while to expand that expression.

Answer 21

Perfect!

Answer 22

hmmmm, i have this next problem y=1/(x-2) here's what i did

Answer 23

For that you can use the chain rule or the quotient rule.

Answer 24

Rewrite as \[y =(x -2)^{-1}\]and perform the same magic.

Answer 25

yea, i started off doing that

Answer 26

y=-(1/(x-2)^2)

Answer 27

Simply express your final answer using positive exponents. Do you want to show me your solution so I can check?

Answer 28

Or you could use the quotient rule:

\[\frac{ (x-2)\frac{ d }{ dx }(1)-(1)\frac{ d }{ dx } (x-2)}{(x-2)^2 }\]

Answer 29

OH! Yes, that is correct!

Answer 30

But what Calculus functions said is a LOT easier.

Answer 31

oh my gosh, i must've done about 15 quotient problems today. i CAN'T STAND THOSE ANYMORE!!!

Answer 32

Haha...

Answer 33

Yes you could use the quotient rule, but it's not necessary.

Answer 34

i got really hard ones involving square roots and such. ruined my friday :(

Answer 35

Chain rule is your friend for square roots.

Answer 36

i'll post a question about the ones i was having trouble with. i still didn't finish 1 problem lol

Answer 37

Go ahead.

Answer 38

Although remember one thing. The only reason quotient rule is not necessary is because the numerator is a constant.

Answer 39

Are you posting them now? Because I have to log off in a few minutes. If I don't see them and you want my help then I'll help the next time I log on. Just message me to remind me.

Answer 40

I can help out for a while if you were to log out...

Answer 41

Do you want to give me one or two quickly right now before I log out?

Answer 42

sure, let me type it out really quick.

Answer 43

differentiate the function:
\[f(x)=\sqrt[3]{x}(\sqrt{x}+3)\]

Answer 44

You have to combine the chain rule and the product rule. Or you can expand it out.

Answer 45

so basically what i've learned is to find the primes and then set it up like f*g' - f' *g / (g)^2

Answer 46

unless i'm using the wrong formula lol

Answer 47

Expanding it will be simpler I believe.

Answer 48

ok gotcha

Answer 49

So that's the same thing as \[x^\frac{ 1 }{ 3 }(x^\frac{ 1 }{ 2 }+3)\]

Answer 50

i wrote it down like this: f prime = 1/3x^-1/6

Answer 51

Which is the same as:

\[x^\frac{ 5 }{ 6 }+3x^\frac{ 1 }{ 2 }\]

Answer 52

oops i meant g prime is x^1/2

Answer 53

Ohh you are using the product rule. It's a LOT easier to expand and use the sum rule instead.

Answer 54

yea, the way my teacher is doing it is just breaking it down into mini steps so we get a basic idea of the rules and such. to see how it all fits in. i do get what you mean though

Answer 55

So the derivative is just:

\[\frac{ 5 }{ 6 }x^\frac{ -1 }{ 6 } +\frac{ 3 }{ 2 }x^\frac{ -1 }{ 2}\]

Answer 56

I used the power rule and the sum rule.

Answer 57

I'm not sure who's answering here. Do you want my explanation or are you good?

Answer 58

Ohh. Sorry @calculusfunctions .

Answer 59

No @Dido525, absolutely no need to apologize. I just wasn't sure if she needs me stick around to help or should I log out.

Answer 60

so that's all we had to do dido? lol. seems like a no brainer

Answer 61

Yeah it's pretty easy.

Answer 62

You can use the product rule, but that was a LOT faster.

Answer 63

i was writing it originally as :

Answer 64

\[(1/3x ^{-1/6})(\sqrt{x}+3)-(\sqrt[3]{x})(x ^{-1/2}) /(\sqrt{x} +3)^2\]

Answer 65

\[y =\sqrt[3]{x}(\sqrt{x}+3)\]
\[y=x ^{\frac{ 1 }{ 3 }}(x ^{\frac{ 1 }{ 2 }}+3)\]
\[y =x ^{\frac{ 5 }{ 6 }}+3x ^{\frac{ 1 }{ 3 }}\]
\[\frac{ dy }{ dx }=\frac{ 5 }{ 6 }x ^{\frac{ -1 }{ 6 }}+\frac{ 1 }{ 3 }x ^{\frac{ -2 }{ 3 }}\]
\[\frac{ dy }{ dx }=\frac{ 5 }{ 6\sqrt[6]{x} }+\frac{ 1 }{ 3\sqrt[3]{x ^{2}} }\]
You can also use the product rule but it's not necessary when the exponent raised to the bracket is just one.

Answer 66

i put a - in x^-1/2 by accident lol. im so tired

Answer 67

@Dido525, I am truly sorry as it was not my intention to offend you or be rude. I sincerely apologize if I came across as snobbish or insensitive. That's not who I am.

Answer 68

@calculusfunctions . You are crazy!!! I never once thought that!! AHAH!!! Lol, you made my day :P .

Answer 69

o.0

Answer 70

@Dido525, good! I'm glad to hear that because I felt terrible. Thanks.

Answer 71

now we can all go to sleep happy :D

Answer 72

@qtexpress101, do you understand now? I'm sure @Dido will do an excellent job at explaining when I log out.

Answer 73

yea, i have a better understanding thanks to the both of you. really appreciate it!

Answer 74

Welcome and Goodnight!

Answer 75

goodnight!