Question

find dy/dx for 4 – xy = y^3.

Answer 1

Can you differentiate y•x?

Answer 2

that'd just be y

Answer 3

You are using the product rule, and the chain rule every time you take derivative of y, and that is you multiply times y'.

Answer 4

What would be the derivative of
\(\color{#000000 }{ \displaystyle x\cdot f(x) }\) ?

Answer 5

Use the product rule for that ...

Answer 6

The derivative of \(f(x)\) is \(f'(x)\)
and the derivative of \(x\) is?

Answer 7

1

Answer 8

for \(x\), and for f(x) ?

Answer 9

no just x

Answer 10

Wait, the derivative of \(x\) with respect to x is 1. Right?

Answer 11

yes

Answer 12

And the derivative of \(f(x)\) is \(f'(x)\) ok?

Answer 13

right

Answer 14

By the product rule
(for differentiable functions of x, F and G)

d/dx [F•G] = F' • G + F • G'

Ok?

Answer 15

yes

Answer 16

d/dx [x•f(x)] = ? (you tell me)

Answer 17

2f(x) ?

Answer 18

h(x) = g(x) * f(x)
h(x) = x * f(x)


so,
g(x) = x
g ' (x) = 1


Using the product rule, we get
h ' (x) = g ' (x) * f(x) + g(x)*f ' (x)


h ' (x) = _________ (fill in the blank)

Answer 19

You seem to struggle doing the product rule...

Answer 20

Can you differentiate \(xe^x\) ?
(Knowing that the derivative of \(e^x\) is \(e^x\) and derivative of x is 1)

Answer 21

Product rule:
d/dx (f•g) = f`g + fg`

Answer 22

Want some examples of product rule?

Answer 23

e^e ?

Answer 24

correct yourself please.

Answer 25

The derivative of \(e^x\) is just \(e^x\).

Answer 26

okay so now what?

Answer 27

Suppose I have the following function:
\(\color{#000000 }{ \displaystyle h(x)=xe^x }\)

Then the derivative of \(h(x)\) would be,

\(\color{#000000 }{ \displaystyle \frac{d}{dx}~xe^x=\left(\frac{d}{dx} x \right) e^x+\left(\frac{d}{dx} e^x \right) x }\)

The derivative of \(x\) (or \(x^1\)) is:
\(\color{blue}{1}\cdot x^{\color{blue}{1}\color{red}{-1}}=1\cdot x^0=1\)
(Knowing that \(y^0=1\) for all y besides y=0)

The derivative of \(e^x\) is just \(e^x\).
(If you like we can go through proving this
via first principles of differentiation).

So, we get;
\(\color{#000000 }{ \displaystyle \frac{d}{dx}~(xe^x)=\left(\frac{d}{dx} x \right) e^x+\left(\frac{d}{dx} e^x \right) x }\)

\(\color{#000000 }{ \displaystyle \frac{d}{dx}~xe^x=\left(1 \right) e^x+\left(e^x \right) x }\)


\(\color{#000000 }{ \displaystyle \frac{d}{dx}~xe^x= e^x+\left(e^x \right) x }\)

Answer 28

If you see something unfamiliar or questionable in what I said, then please ask.

Answer 29

(take your time tho)

Answer 30

I think I understand but how do I use this to solve the problem?

Answer 31

Please note:\[4-xy=y^3\]

Answer 32

is an implicit function. Theoretically, it could be solved for y, but that'd be hard and not worth the trouble.

Answer 33

Did you get the product rule?
Do you know the power rule (what I applied to differentiate the x)?
Do you know the chain rule?

Answer 34

So you must jump in and take the derivative of each term with respect to x.

Answer 35

In other words, evaluate\[\frac{ d }{ dx}(4-xy=y^3)\]

Answer 36

what is \[\frac{ d }{ dx}4?\]

Answer 37

Hint: the derivative of a constant is always ___________

Answer 38

zero

Answer 39

Right. Now let's move on to

Answer 40

\[\frac{ d }{ dx }[-xy]\]

Answer 41

You could simplify matters by taking the neg. sign out.

Answer 42

-y

Answer 43

\[-\frac{ d }{ dx }xy\]

Answer 44

or do you want y?

Answer 45

solomon has gone over that with you already; please go back for a quick look at the product rule, and then at the derivative of xy with respect to x.

Answer 46

Again: xy is a product, and we want the derivative of this product, so "do you want y?" is irrelevant.

Answer 47

So, writing everything out formally, the derivative of xy is

Answer 48

...I'll let Solomon type that for us.

Answer 49

no i'm say the derivative of xy is y right?

Answer 50

Looks like there is a little lack of prerequisite knowledge... but we can fix that if you bear with us.... it might take time, but wouldn't you agree it is worth it?

Answer 51

Do you know the Power Rule?
If you do please differentiate \(x^3\) for me.

Answer 52

Trisa: Sorry, no.
Solomon: Please stick to the derivative of xy first, then worry about the power rule later.

Answer 53

Formally, the derivative with respect to x of xy is as follows:\[\frac{ d }{ dx }xy=x \frac{ dy }{ dx }+y \frac{ dx }{ dx }\]

Answer 54

This is the applicaTION OF THE product rule for differentiation to xy. Note that we are assuming that y is a function of x. That's why I had to say NO to your question earlier.

Answer 55

So, Tri-: can you simplify dx/dx?

Answer 56

1

Answer 57

Good!
So, what do you have left?

Answer 58

just y

Answer 59

so now I simplify dy/dx?

Answer 60

Yes.\[\frac{ d }{ dx }xy=x \frac{ dy }{ dx }+y \]

Answer 61

d/dx xy = x y/2 +y

Answer 62

No. Please move on to the last term, which is y^3. What is the derivative of that?

Answer 63

I'm sorry, but x y/2 + y is not correct. Please compare it to my last previous equation, above.

Answer 64

I'm getting the results I have by applying the product rule to xy.

Answer 65

Try once more.

The middle term is -xy. What is the derivative, with respect to x, of that?

Answer 66

didn't u just tell me what the derivative of xy was just now??

Answer 67

Yes. I want to be sure you agree with that before we move on.

Answer 68

\[\frac{ d }{ dx }xy=x \frac{ dy }{ dx }+y\]

Answer 69

x d/dx + y dx/dx

Answer 70

Now enclose that in parentheses and write a negative sign in front. Then you'll have the derivative of -xy with respect to x, and assuming that y is a function of x.

Answer 71

alright but how is dy/dx not simplified to y/x

Answer 72

\[\frac{ d }{ dx }[-xy]=-(x \frac{ dy }{ dx }+y)\]

Answer 73

I'm glad you asked that question. dy/dx cannot in any way be reduced.

d is not a number, but an operator (command).

dy/dx as a whole reads "the derivative of y with respect to x."

Answer 74

ha nvm lets just continue

Answer 75

so now I find the derivative of y^3?

Answer 76

The last term we have to deal with is y^3.
Find the derivative of that with respect to x. Be certain to apply the power rule AND the chain rule.

Answer 77

3y^2

Answer 78

Great, but you haven't yet applied the chain rule. Remember, y is assumed to be a function of x, and thus you must differentiate y with respect to x.

Answer 79

So, the derivative of y^3 with respect to x is 3y^2 times what?

Answer 80

sooo zero?

Answer 81

The derivative of y with respect to x is denoted by dy/dx. You will see this again and again.

Write the whole derivative of y^3 with respect to x here:

Answer 82

Everything you need to do that is included in the most recent 3 entries, above.

Answer 83

compare this to

Answer 84

\[3y^2\frac{ dy }{ dx }\]

Answer 85

sorry it's hard for you to understand this right now, but dy/dx is NOT equal to 1.
Instead, dy/dx is a LABEL (like your name is a label) for "the derivative of y with respect to x."

Answer 86

To get 3y^2, we used the power rule.
We remember that y is assumed to be a function of x, so we follow the chain rule and write dy/dx immediately after 3y^2.

Answer 87

In summary:\[\frac{ d }{ dx }4=0\]

Answer 88

and ...

Answer 89

\[-\frac{ d }{ dx }xy =-[x \frac{ dy }{ dx }+y]\]

Answer 90

... and ...\[\frac{ d }{ dx }y^3=3y^2\frac{ dy }{ dx }\]

Answer 91

I'm afraid you'll have to take my word for this for the time being. You can ask all the questions you want, perhaps tomorrow.

Answer 92

So, put all 3 of these derivatives together to obtain the derivative of the implicit function 4-xy=y^3.

Answer 93

Could you do that now, please? Don't bother with \[\frac{ d }{ dx}4,\]

Answer 94

because it's zero. What's left?

Answer 95

-[x dy/dx+ y] = 3y^2 dy/dx

Answer 96

You are now writing an equation with the derivative of -xy on the left of the = sign and the deriv. of y^3 on the right. Yes, y ou have that correct.

Answer 97

The last step of the process of finding the derivative is to SOLVE FOR dy/dx. Once again, dy/dx reads "the derivative of y with respect to x" and is not 1 and is not 0.

Answer 98

There are 3 terms in the equation you've just written. 2 have dy/dx in them, and 1 does not. Can you agree with that?