Question

Find dy/dx by implicit differentiation: 4 cos x sin y = 1

Answer 1

Hello, Jay!
Let's take a look at 4 cos x sin y = 1. Do you recognize that (cos x)(sin y) is a product?

Answer 2

yes sir

Answer 3

Jay, supposing you had (cos x)(sin x). How would you go about finding the derivative of that with respect to x? (This is for practice.) Can you comfortably apply the product rule?

Answer 4

Which form of the product rule are you most comfortable with ?

Answer 5

product rule is the easy to solve.
but my classmate told me that he used the implicit rule
sorry for my grammar i'm not fluent in english

Answer 6

i think product rule is the easy way to solve

Answer 7

Actually, we do have to use implicit differentiation, and are going to use it, soon.

For now, please type out the form of the product rule with which you're most familiar.

Answer 8

i have no idea.
because i'm such a noob

Answer 9

it's my assignment. pls help me :(

Answer 10

-4sin(x)*sin(y) + 4cos(x)*cos(y)*dy/dx = 0

Answer 11

Let me do a quick review for you:

The product rule looks like this:

Answer 12

\[\frac{ d }{ dx }u*v = u \frac{ dv }{ dx }+v \frac{ du }{ dx }\]

Answer 13

Here we are assuming that both u and v are functions of x.
Have you seen this formula before? Have you used it before?

Answer 14

Jay?

Answer 15

wait im thinking

Answer 16

Please write that formula down, Jay. You'll need it again and again in Calculus.
Label it "product rule"

Answer 17

Another form of the product rule is (u*v) ' = u*v " + v*u '

Answer 18

d/dx [f(x)g(x)] = f(x)g'(x) + g(x)f'(x)

Answer 19

That's fine. We're on the same wavelength. Very good.

Answer 20

Now, back to the example problem I gave you. y=(sin x) (cos x) is a product. We'll apply the product rule to find the derivative of this function.

Answer 21

Let f(x) = sin x
Let g(x) = cos x
What are f '(x) and g '(x) ?

Answer 22

i dont know how to solve that :(

Answer 23

Is this a calculus course, or are you learning some calculus in some other course, such as Advanced Placement Math?

Answer 24

sin x d/dx (cos x) + cos x d/dx (sin x) ?

Answer 25

Jay, I need your complete attention if you want my help in learning this material. OpenStudy says you are "just looking around." Are you willing and able to give me your complete attention?

Answer 26

yes i will give it to you

Answer 27

sorry for that

Answer 28

If f(x) = sin x, the derivative if f '(x) = cos x. (You really do have to know that by heart.)

Answer 29

If g(x) = cox x, the derivative is g ' (x) = - sin x.

I'd suggest you write all of this down for later reference, as you will see these functions, formulas and properties again and again.

Answer 30

okay i get it

Answer 31

Again, let's focus on finding the derivative of y= (sin x)(cos x).
f(x) * g(x)

You now know what f(x), g(x), f '(x) and g '(x) are.

Can you apply the product rule to finding the derivative dy/dx, also using the derivatives of sin x and cos x that I've given you?

Answer 32

wait give me a minute

Answer 33

Write the derivative in this form:\[y ' = f * g ' + g * f '\]

Answer 34

0 = (sin x ) (-sin x) + ( cos x ) ( cos x ) ?

Answer 35

y ' is not zero here, but the rest of your expression is fine. Why did you think y ' = 0 ?

Answer 36

Remember, this is a practice problem; it is not the actual problem that you need to solve.

Answer 37

d/dx y
i don't know sorry

Answer 38

Actually, Jay, we are not using the power rule. We need and are using the product rule.
Let me summarize this again:
If y = sin x cos x
then f = sin x , g = cos x, f ' = cos x and g ' = ??? (can you finish this by telling me what g ' is equal to when g = cos x?

Answer 39

g ' = ?

Answer 40

- sin x

Answer 41

Great! Progress!

Answer 42

So, Jay, if y = sin x cos x,
the derivative is
y ' = ( )( ) + ( ) ( )
please fill in the blanks. (This should not be new to you any longer, since we've done this already this morning.)

Answer 43

(sin x) (-sin x) + (cos x) ( cos x)

Answer 44

Very nice! Please label that. Your label should be y ' .

Type out y ' and then copy and paste your previous result.

Answer 45

y ' = (sin x) (-sin x) + (cos x) (cos x)

Answer 46

Very nice. I keep asking you to type things out so that you can get the practice you need in doing so.

Answer 47

thanks :)

Answer 48

Now suppose you have y = 4 [ sin x cos x ]. The only difference here is that you've multiplied sin x cos x by 4. What will the derivative of y = 4 [ sin x cos x] be?

Answer 49

Hint: Use the "Constant Coefficient Rule for derivatives" here.

Answer 50

Don't sweat it; this is one of the easiest rules of differentiation to apply.

Answer 51

y' = 0

Answer 52

No, because that 4 is a constant coefficient.

The derivative of 4 is zero, yes, but the derivative of 4x is definitely not zero.

If y = 4 [ sinx cos x ], the derivative y ' is y ' = 4 [ ? ]

Answer 53

Don't sweat this; you already have the answer (above); just copy and paste it within the brackets [ ]

Answer 54

y' = 4 [ sinx * -sinx + cos x * cos x ]

Answer 55

That's great. You're doing fine.

Answer 56

Review:

If y = sin x, (dy/dx) or y ' = cos x, right?

Answer 57

yup

Answer 58

Now suppose we throw out that 'x' and replace it with 'u' and note that 'u' is itself a function.

The chain rule requires us to find the derivative of that 'u' with respect to the independent variable, x.

Answer 59

Do you recall having heard or read about the chain rule as it applies here?

Answer 60

i have no idea about in chain rule :3

Answer 61

For now, you need to realize that 'u' is now a function in its own right, and that if we want to find the derivative of y = sin u, we must first differentiate with respect to u and then differentiate u itself with respect to x.

Here's what this adds up to:

Answer 62

Basic function y = sin x
Derivative of y = sin x is y ' = cos x. Done. Finished. That's it.

More complicated case: y = sin u, where 'u' is a separate function.

Answer 63

Goal: find the derivative of y = sin u with respect to x.

Method: First, find the derivative of sin u with respect to u: As you'll recognize from earlier, y ' = cos u.

Answer 64

To finish differentiating y = sin u, you must now find the derivative of the function u and multiply it by cos u:

If y = sin u , then y ' = cos u * u '

That's it.

Example;

Answer 65

fIND THE Derivative of y = sin (2x).

nOTE THAT u = 2x here, and note that u is a separate function of x whose derivative is

u ' = 2. OK with that?

Answer 66

then y ' = cos 2x * 2 ?

Answer 67

Perfect. Jay, you know your stuff!

Answer 68

Now we're going to use implicit differentiation to find y ' if 4 cos x sin y = 1. This is the problem you posted originally.

Answer 69

yeah. and i have no idea where i start

Answer 70

I'll bet you do too have an idea of what to do!

Answer 71

Write this out to show yourself and me what you are supposed to do:

"Find y ' if 4 cos x sin y = 1 "

Start by enclosing the left side and then the right side within parentheses:

(4 cos x sin y ) = (1)

Answer 72

Now indicate that you are to take the derivative of both sides:

(4 cos x sin y ) ' = (1) '

Answer 73

First, let's look at the right side. What is the derivative of a constant?

Answer 74

Equivalently, what is the slope of a horizontal straight line?

Answer 75

i don't how to get the slope of that line

Answer 76

Review: the slope of a horizontal line is zero.

The slope of the horizontal line y = 1 is zero.

The slope of a line is equal to the derivative of the linear function y = 1, which again is zero.

What is the derivative of y = 1?

This is [ 1 ] '.

Answer 77

I am saying the same thing in several different ways:

The slope of a horizontal line is zero.
y = 1 represents a horizontal line. Therefore, the derivative of y is zero because the derivative of 1 is zero; the derivative of a constant is zero.

Answer 78

What is the right side of this equation equal to? (4 cos x sin y ) ' = (1) '

Answer 79

(1) ' = 0

Answer 80

still there?

Answer 81

Yes

Answer 82

I'm sorry OpenStudy froze up.
You have the correct result for the right side of y our equation: it is zero (0).

Answer 83

Therefore, (4 cos x sin y ) ' = (1) '

becomes (4 cos x sin y ) ' = 0.

Answer 84

how about on the left side what should i do

Answer 85

Remember, that '4' is a constant coefficient, so you can re-write

(4 cos x sin y ) ' = 0 as (4) * (cos x sin y ) ' = 0.

We have already found the derivative with respect to x of (cos x sin y). Could you review our work and find that derivative, and then plug it into (4)* ( ) = 0 ?

Answer 86

Jay, I'm glad you're interested in communicating with me ( e. g., through Facebook), but please let's finish the solution to this problem, OK?
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Answer 87

4 [ -sin x sin y + cos x cox y ] = 0 ?

Answer 88

then whats next?

Answer 89

Notice that I deleted my last post; I'd made a mistake there.

It'll take just a minute: Let's AGAIN find the derivative of cos x sin y with respect to x:

Applying both the product and chain rules,

the derivative of cos x sin y with respect to x is :

(cos x) (-cos y * y ') + (sin y ) (-sin x)

Check that over, please, and see whether you agree or disagree or are not sure.

Answer 90

I had to take the derivative of y because we are assuming that y is a separate function of x.

Answer 91

but why it become (cos x ) (-cos y * y' ) + ( sin y ) (-sin x) ?

Answer 92

To answer that, I'd need to know what you understand and what you don't understand.

What's new about this problem, compared to my examples, is that 'y' is now a separate function of x, requiring us to use the Chain Rule.

Answer 93

Point out what you do understand and what y ou do not understand about

(cos x ) (-cos y * y' ) + ( sin y ) (-sin x)

Answer 94

the - cos y * y'

Answer 95

why it becomes negative?

Answer 96

Rather than answer that, I'm asking you to show me what you think the correct result should be.

Answer 97

( cos x ) (cos y * y ') + ( -sin x ) ( sin y)

Answer 98

I agree with your ( cos x ) (cos y * y ') and admit I must have made a mistake there. Glad you've found it.