- #1

- 457

- 0

here's what i did

d/dx 4 cosx sin y)= 1 d/dx

4sinx cosy=0

y'= 4 sinx cosy

but the answer is tan x tan y

- Thread starter afcwestwarrior
- Start date

- #1

- 457

- 0

here's what i did

d/dx 4 cosx sin y)= 1 d/dx

4sinx cosy=0

y'= 4 sinx cosy

but the answer is tan x tan y

- #2

- 457

- 0

do ihave to use the power rule and the chain rule in this problem

- #3

- 457

- 0

well what do i have to do may u show me the steps or correct me,

- #4

hage567

Homework Helper

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- #5

- 457

- 0

ok ill try it out

- #6

- 457

- 0

is this right or am i wrong with the product rule of this

- #7

- 457

- 0

then i got 4 sin X + sin x * cos y dy/dx

- #8

- 457

- 0

wheres the help man

- #9

hage567

Homework Helper

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No, this doesn't look right. If you are going to include the 4 in the product rule, you will have to apply it twice since that way you technically have three terms.

is this right or am i wrong with the product rule of this

In the second term why did the cosx turn into a sinx? You already took the derivative of that in the first term, so the rule says you leave it alone in the second.then i got 4 sin X + sin x * cos y dy/dx

OK

- #10

- 457

- 0

so like this ( d/dx cos X + sin X d/dx) sin y dy/dx

- #11

- 457

- 0

I mean 4 sin X + cos X * siny dy/dx

- #12

- 457

- 0

then i change it to sin y dy/dx= -4 sin x / cos x

- #13

- 457

- 0

sin y dy/dx= -4cos - sin x

- #14

- 457

- 0

so 4 = 0 right because the rule says d/dx (C) =0

- #15

- 457

- 0

so then this is sin y dy/dx= cos - sin x

then its dy/dx= cos- sin x/ sin y

then its dy/dx= cos- sin x/ sin y

- #16

- 457

- 0

do i have to use the quotient rule on this part

- #17

- 572

- 3

I think that you are misunderstanding the product rule. Firstly, since 4 is a constant, you can take it out of the situation. So

[tex]\frac{d}{dx}4cosxsiny[/tex] = [tex] 4 \frac{d}{dx}cosxsiny[/tex]

Now, the product rule is [tex] \frac{d}{dx}[f(x)g(x)]=f(x)g'(x) + g(x)f'(x) [/tex] I can demonstrate the proof if you wish but I won't clutter up the post unnecesarily.

__I am still editing this, but I see that you are still posting.. PLEASE use the edit button..... I am still editing this post so check back in a few minutes...__

Okay, anyways... So I will do an example problem and show you how to do it.

Let me explain implicit differentiation for you. When you are differentiating terms involving x alone, you can differentiate as usual. However, when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x.

Lets say the problem was:

find dy/dx given that [tex]y^{3}+y^{2}-5y-x^{2}=-4[/tex]

Solution:

1.Differentiate both sides of the equation with respect to x.

[tex] \frac{d}{dx}[y^{3}+y^{2}-5y-x^{2}]=\frac{d}{dx}[-4] [/tex]

[tex] \frac{d}{dx}[y^{3}] + \frac{d}{dx}[y^{2}] - \frac{d}{dx}[5y] - \frac{d}{dx}[x^{2}] = \frac{d}{dx}[-4][/tex]

[tex]3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x=0[/tex] (chain rule)

2. Collect the dy/dx terms on the left side of the equation.

[tex]3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}=2x[/tex]

3.Factor dy/dx out of the left side of the equation

[tex]\frac{dy}{dx}(3y^{2}+2y-5)=2x[/tex]

4.Solve for dy/dx by dividing by [tex]3y^{2}+2y-5[/tex]

[tex]\frac{dy}{dx}=\frac{2x}{3y^{2}+2y-5}[/tex]

There ya go... Do you understand now?

[tex]\frac{d}{dx}4cosxsiny[/tex] = [tex] 4 \frac{d}{dx}cosxsiny[/tex]

Now, the product rule is [tex] \frac{d}{dx}[f(x)g(x)]=f(x)g'(x) + g(x)f'(x) [/tex] I can demonstrate the proof if you wish but I won't clutter up the post unnecesarily.

Okay, anyways... So I will do an example problem and show you how to do it.

Let me explain implicit differentiation for you. When you are differentiating terms involving x alone, you can differentiate as usual. However, when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x.

Lets say the problem was:

find dy/dx given that [tex]y^{3}+y^{2}-5y-x^{2}=-4[/tex]

Solution:

1.Differentiate both sides of the equation with respect to x.

[tex] \frac{d}{dx}[y^{3}+y^{2}-5y-x^{2}]=\frac{d}{dx}[-4] [/tex]

[tex] \frac{d}{dx}[y^{3}] + \frac{d}{dx}[y^{2}] - \frac{d}{dx}[5y] - \frac{d}{dx}[x^{2}] = \frac{d}{dx}[-4][/tex]

[tex]3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x=0[/tex] (chain rule)

2. Collect the dy/dx terms on the left side of the equation.

[tex]3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}=2x[/tex]

3.Factor dy/dx out of the left side of the equation

[tex]\frac{dy}{dx}(3y^{2}+2y-5)=2x[/tex]

4.Solve for dy/dx by dividing by [tex]3y^{2}+2y-5[/tex]

[tex]\frac{dy}{dx}=\frac{2x}{3y^{2}+2y-5}[/tex]

There ya go... Do you understand now?

Last edited:

- #18

- 457

- 0

ok can u demonstrate it too me

- #19

- 457

- 0

which equals cos x (cos y dy/dx) + siny d /dx (sin X)

- #20

- 572

- 3

Ok, I don't really see where you are getting the dy/dx in the first post... Can you explain it to me? You have d/dx in front of siny, so you haven't differentiated it yet, why do you have dy/dx there?

Last edited:

- #21

- 457

- 0

then i did this - cos x/ sin y - sin x/ cos y = tan x tan y

- #22

- 457

- 0

i understand that ok so i do thisI think that you are misunderstanding the product rule. Firstly, since 4 is a constant, you can take it out of the situation. So

[tex]\frac{d}{dx}4cosxsiny[/tex] = [/tex] 4 \frac{d}{dx}cosxsiny[/tex]

Now, the product rule is [tex] \frac{d}{dx}[f(x)g(x)]=f(x)g'(x) + g(x)f'(x) [/tex] I can demonstrate the proof if you wish but I won't clutter up the post unnecesarily.

I am still editing this, but I see that you are still posting.. PLEASE use the edit button..... I am still editing this post so check back in a few minutes...

Okay, anyways... So I will do an example problem and show you how to do it.

Let me explain implicit differentiation for you. When you are differentiating terms involving x alone, you can differentiate as usual. However, when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x.

Lets say the problem was:

find [tex] \frac{dy}{dx}[/tex] given that [tex]y^{3}+y^{2}-5^{y}-x^{2}=-4[/tex]

Solution:

1.Differentiate both sides of the equation with respect to x.

[tex] \frac{d}{dx}[y^{3}+y^{2}-5y-x^{2}]=\frac{d}{dx}[-4] [/tex]

[tex] \frac{d}{dx}[y^{3}] + \frac{d}{dx}[y^{2}] - \frac{d}{dx}[5y] - \frac{d}{dx}[x^{2}] = \frac{d}{dx}[-4][/tex]

[tex]3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x=0[/tex] (chain rule)

2. Collect the dy/dx terms on the left side of the equation.

[tex]3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}=2x[/tex]

3.Factor dy/dx out of the left side of the equation

[tex]\frac{dy}{dx}(3y^{2}+2y-5)=2x[/tex]

4.Solve for dy/dx by dividing by [tex]3y^{2}+2y-5[/tex]

[tex]\frac{dy}{dx}=\frac{2x}{3y^{2}+2y-5}[/tex]

There ya go... Do you understand now?

- #23

- 457

- 0

d/dx 4 * d/dx cos X * d/dx sin y = d/dx (1)

d/dx 0 * d/dx sin x * d/dx cos y =0

then d/dx( sinx * cosy= 0

- #24

- 1,860

- 0

You violated a ton of mathematical rules here, like getting multiplication out of addition.then i did this - cos x/ sin y - sin x/ cos y = tan x tan y

Remember the product rule is [tex]\frac{d}{dx}u*v = u \frac{d}{dx}v + v \frac{d}{dx}u = uv' + vu'[/tex]

Now just apply this to your implicit differentiation problem where u = cosx and v = siny.

- #25

- 572

- 3

Firstly, d/dx (4 cos x * sin y)= 1 is not true... (why?)

d/dx 4 * d/dx cos X * d/dx sin y = d/dx (1)

d/dx 0 * d/dx sin x * d/dx cos y =0

then d/dx( sinx * cosy= 0

Secondly, d/dx 4 * d/dx cos X * d/dx sin y = d/dx (1), you are not applying the product rule correctly. Nor the constant multiple rule.

I gotta go but if you want to survive, you have to review... lots. Do you want me to do an example with the product rule before I go?

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