# Find dy/dx by implicit differetiation HELP IMMEDIATLY

• afcwestwarrior

#### afcwestwarrior

of 4 cos x sin y =1
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

## Answers and Replies

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

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

I think you need to use the product rule since it is the product of the two functions (cosx and siny).

ok ill try it out

here's what i did 4 d/dx (Cos x) + cos x * d/dx (4) * sin y dy/ dx

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

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

wheres the help man

here's what i did 4 d/dx (Cos x) + cos x * d/dx (4) * sin y dy/ dx

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

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.

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

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.

OK make sure you understand the product rule. Don't concentrate on the 4, it is just a constant. You could just take it out for now. Try breaking it up like (cosx)(siny) as your two terms. Now try it again.

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

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

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

wait a minute so 4 is left alone, so it's like this cos d/dx (x) + (x) d/dx (cos) * siny dy/dx then i change it too

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

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

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

do i have to use the quotient rule on this part

I think that you are misunderstanding the product rule. Firstly, since 4 is a constant, you can take it out of the situation. So
$$\frac{d}{dx}4cosxsiny$$ = $$4 \frac{d}{dx}cosxsiny$$

Now, the product rule is $$\frac{d}{dx}[f(x)g(x)]=f(x)g'(x) + g(x)f'(x)$$ 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 $$y^{3}+y^{2}-5y-x^{2}=-4$$

Solution:

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

$$\frac{d}{dx}[y^{3}+y^{2}-5y-x^{2}]=\frac{d}{dx}[-4]$$
$$\frac{d}{dx}[y^{3}] + \frac{d}{dx}[y^{2}] - \frac{d}{dx}[5y] - \frac{d}{dx}[x^{2}] = \frac{d}{dx}[-4]$$
$$3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x=0$$ (chain rule)

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

$$3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}=2x$$

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

$$\frac{dy}{dx}(3y^{2}+2y-5)=2x$$

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

$$\frac{dy}{dx}=\frac{2x}{3y^{2}+2y-5}$$

There you go... Do you understand now?

Last edited:
ok can u demonstrate it too me

is it like this cos x d/dx (sin y) dy/dx + siny d/dx cos X

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

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

I think that you are misunderstanding the product rule. Firstly, since 4 is a constant, you can take it out of the situation. So
$$\frac{d}{dx}4cosxsiny$$ = [/tex] 4 \frac{d}{dx}cosxsiny[/tex]

Now, the product rule is $$\frac{d}{dx}[f(x)g(x)]=f(x)g'(x) + g(x)f'(x)$$ 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 $$\frac{dy}{dx}$$ given that $$y^{3}+y^{2}-5^{y}-x^{2}=-4$$

Solution:

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

$$\frac{d}{dx}[y^{3}+y^{2}-5y-x^{2}]=\frac{d}{dx}[-4]$$
$$\frac{d}{dx}[y^{3}] + \frac{d}{dx}[y^{2}] - \frac{d}{dx}[5y] - \frac{d}{dx}[x^{2}] = \frac{d}{dx}[-4]$$
$$3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x=0$$ (chain rule)

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

$$3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}=2x$$

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

$$\frac{dy}{dx}(3y^{2}+2y-5)=2x$$

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

$$\frac{dy}{dx}=\frac{2x}{3y^{2}+2y-5}$$

There you go... Do you understand now?

i understand that ok so i do this

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

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

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

You violated a ton of mathematical rules here, like getting multiplication out of addition.

Remember the product rule is $$\frac{d}{dx}u*v = u \frac{d}{dx}v + v \frac{d}{dx}u = uv' + vu'$$

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

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

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

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

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 got to 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?

cos x (cos y') + sin y (sin X)

or dy/dx (cos x*cos y' + siny * sin X=0

Lol, what are you going on about? I am so lost, is this a joke?

is this right

Lol, what are you going on about? I am so lost, is this a joke?

ill show u what i did
4 cos sin y =1

d/dx 4 * cos x d/dx sin y + sin y d/dx cos x = d/dx 1

d/dx(cos x * cos y' + sin y sin x)=0

You are really struggling with the product rule, but look at it as substitution.

u = cosx and v = siny

d/dx u v = uv' + vu'

so

d/dx cosx*siny = cosx*d/dx(siny) + siny*d/dx(cosx)

Try showing what the derivative of d/dx(x*cosx) is.

isnt the derivative of cos sin and sin cos

so does it look like this cos x/ cosy + siny /sin x

the bottoms are derivatives

In words, this means take the derivative of sine with respect to y (where the implicit differentiation comes in) and multiply it by cosine, then add the derivative of cosine multiplied by sine.

You keep throwing out equations without having any mathematical reason for getting there, which will definitely not get you to the answer. Now, I understand you are desperate, but you got to stop flailing and just think about it. I have given you everything you need, the rest is up to you.

Okay, do you KNOW what the product rule and chain rule are?

Do you know what the derivative of sin(x)cos(x) is?

Do you know how to find the derivative of y if x+ y= 1 and y is a function of x?

In other words, let's find out what you do know before we start making suggestions!

(Oh, and the answer to
afcwestwarrior said:
isnt the derivative of cos sin and sin cos
is "NO!")

I strongly recommend that you go back and review your basic differentiation laws.

Last edited by a moderator: