Question

Find the derivative of the function: √-9-5x.
dy/dx=? help much appreciated :)

Answer 1

Is that\[y =\sqrt{-9-5x}\]

Answer 2

You can use the Chain Rule. I'm sure Sir @calculusfunctions will tell you more.

Answer 3

@calculusfunctions, yeah

Answer 4

\[\rm {d \over dx}f(g(x)) = g'(x)f'(g(x))\]

Answer 5

OK then as @ParthKohli said very accurately, do you know the chain rule or the power of a function rule?

Answer 6

^ lol

Answer 7

@ParthKohli got it

Answer 8

Yes, that's another one you can use! Though you have to use Chain Rule, and the power rule comes in that.

Answer 9

Yes @ParthKohli that is excellent by there is an easier rule which is also a form of the chain rule.

Answer 10

@frankypoo Give me a few minutes to type up a explanation for you.

Answer 11

Alright (:

Answer 12

He's a great teacher, isn't he?

Answer 13

He knows what he's doing aha, I'm doing calc atm, It's all in his name aha

Answer 14

doing Derivatives and with all of the rules and types of problems I get mixed up :c

Answer 15

Chain Rule (Power of a Function Rule):\[y =C[(f(x)]^{n}\]where C and n are any real number.
\[\frac{ dy }{ dx }=Cn[f(x)]^{n -1}f \prime(x)\]
Example:\[y =5\sqrt[5]{x ^{2}+2}\]First rewrite\[y =5(x ^{2}+2)^{\frac{ 1 }{ 5 }}\]Now\[\frac{ dy }{ dx }=(5)(\frac{ 1 }{ 5 })(x ^{2}+2)^{\frac{ 1 }{ 5 }-1}(\frac{ d(x ^{2}+2) }{ dx })\]
\[\frac{ dy }{ dx }=(x ^{2}+2)^{\frac{ -1 }{ 4 }}(2x)\]
\[\frac{ dy }{ dx }=\frac{ 2x }{ \sqrt[4]{x ^{2}+2} }\]

Answer 16

@frankypoo if you understand then try your question now. @ParthKohli thank you and have you seen this form of the chain rule before? It is still the Chain rule but it's called the Power of a function rule.

Answer 17

No, I've not... I'm currently learning about it. Thank you sir!

Answer 18

Of course @ParthKohli Nice profile picture.

Answer 19

Thanks sir, and yours too! (I love Sheldon Cooper :) )

Answer 20

@frankypoo where are you?

Answer 21

here ahaha.

Answer 22

"scooby doo, where are you?!" just reminded me of that

Answer 23

I'm about to try it, give me a sec.

Answer 24

Yes lol

Answer 25

Take your time. I'm just humming the theme song, accept I replace Scooby Doo with @frankypoo in the lyrics. LOL

Answer 26

ahaha, alright

Answer 27

lol

Answer 28

I'm still confused :s

Answer 29

With which part?

Answer 30

EVERYTHING

Answer 31

LOL

Answer 32

I think I need another example.

Answer 33

OK we'll take it one step at a time. Babypoo steps! LOL but first I need to know if you at least understand the the rule? If not that's OK too because you will once I'm through.

Answer 34

unless the answer is... \[\frac{ dy }{ dx }=\frac{ x }{ \sqrt{-5x-9}}\]

Answer 35

NO!

Answer 36

which I doubt

Answer 37

LOL

Answer 38

OK step 1: Can you first rewrite the function without radicals exactly like I did in the example?

Answer 39

y=(-9-5x)

Answer 40

No...What are you missing?

Answer 41

.-.

Answer 42

\[\sqrt{x}=x ^{\frac{ 1 }{ 2 }}\]using this fact, try again!

Answer 43

dafuq

Answer 44

Do you know the rational exponent law?

Answer 45

noo,

Answer 46

ill search it real quick.

Answer 47

No, I'll teach it to you @frankypoo

Answer 48

Teach it quick-like

Answer 49

A rational power, m/n, raised to the base x is equal to the nth root of x raised to the mth power or the mth power of the nth root of x. In other words \[x ^{\frac{ m }{ n }}=(\sqrt[n]{x})^{m}=\sqrt[n]{x ^{m}}\]

Example:\[8^{\frac{ 2 }{ 3 }}=(\sqrt[3]{8})^{2}=2^{2}=4\]

Answer 50

makes sense

Answer 51

Please do not rush me when I am teaching in this format. I type slowly with care to make sure that no errors are made in the lessons I give you. Alright?

Answer 52

OK so then if it makes sense then can you rewrite your function without radicals again?

Answer 53

@frankypoo I had to go half an hour earlier but I want to finish teaching you first. My point is that I didn't rush you to learn faster so then perhaps you should extend the same courtesy.

Answer 54

Do you need an example of how to use the rational exponent law to rewrite the function without radicals?

Answer 55

would be appreciated

Answer 56

Suppose \[y =\sqrt{3x ^{5}-x +4}\] Then by the rational exponent law \[y =(3x ^{5}-x +4)^{\frac{ 1 }{ 2 }}\]
Understand? Now can you rewrite yours?

Answer 57

where did the 1/2 come from?

Answer 58

I thought you said you understood the rational power law. Please take another look while I explain again.

Answer 59

y=(-5x-9)^1/2

Answer 60

Excellent!!! Understand now??

Answer 61

If you have a nth root of a function f(x) raised to the exponent m, then the function f(x) is raised to the exponent m/n.

Answer 62

yea, bit more sense

Answer 63

\[y =(-9-5x)^{\frac{ 1 }{ 2 }}\]According to what I demonstrated, what is f(x)?

Answer 64

What is the exponent and what is the number in front of the bracket?

Answer 65

Can you please tell me at least that much?

Answer 66

I'm trying

Answer 67

Baby steps remember? All I asked is if\[y =(-9-5x)^{\frac{ 1 }{ 2 }}\]then what is the number in front of the brackets and what is the exponent?

Answer 68

There is no number visible in front of the bracket so it's a 1. What is the exponent?

Answer 69

in front of the brackets there's an invisible 1, exponent 1/2

Answer 70

Very good! Now you multiply those numbers so what do you get?

Answer 71

What is 1 multiplied by 1/2?

Answer 72

1/2.

Answer 73

Excellent! Now if you raise the function to an exponent one less than what it is now, what do you get?

Answer 74

(1/2) - 1 = ?

Answer 75

-1/2

Answer 76

Correct! Now finally if you take the derivative of the function inside the bracket, what do you get?

Answer 77

-5

Answer 78

What is the derivative, dy/dx, if y = -9 - 5x ?

Answer 79

Right!

Answer 80

\[\frac{ -5 }{ \sqrt{-5x-9} }\]

Answer 81

what am I missing?

Answer 82

Almost! What's missing? Where is the 1/2?

Answer 83

isnt it 2 by the radical?

Answer 84

And please write it properly as dy/dx = ?

Answer 85

That's what I meant by where is the 1/2 ?

Answer 86

So can you please write your final answer again with dy/dx?

Answer 87

\[\frac{ dy }{ dx }=\frac{ -5 }{ \sqrt[2]{-5x-9} }\]

Answer 88

NO! It should look like this\[\frac{ dy }{ dx }=\frac{ -5 }{ 2\sqrt{-5x -9} }\]You have your 2 in the wrong spot!!

Answer 89

Understand now?

Answer 90

I meant to do that :D

Answer 91

yessir, thanks for your help and your patience.

Answer 92

Very Good! @frankypoo congratulations on finally solving another mystery just like Scooby Doo.

Answer 93

hah, alright!:)

Answer 94

You're welcome! Anytime.

Answer 95

I don't even get a medal? LOL

Answer 96

there ya go :)