# Find dy/dx by implicit differetiation HELP IMMEDIATLY

1. Mar 6, 2007

### 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

2. Mar 6, 2007

### afcwestwarrior

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

3. Mar 6, 2007

### afcwestwarrior

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

4. Mar 6, 2007

### hage567

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

5. Mar 6, 2007

### afcwestwarrior

ok ill try it out

6. Mar 6, 2007

### afcwestwarrior

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

7. Mar 6, 2007

### afcwestwarrior

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

8. Mar 6, 2007

### afcwestwarrior

wheres the help man

9. Mar 6, 2007

### hage567

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.

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.

10. Mar 6, 2007

### afcwestwarrior

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

11. Mar 6, 2007

### afcwestwarrior

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

12. Mar 6, 2007

### afcwestwarrior

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

13. Mar 6, 2007

### afcwestwarrior

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

14. Mar 6, 2007

### afcwestwarrior

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

15. Mar 6, 2007

### afcwestwarrior

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

16. Mar 6, 2007

### afcwestwarrior

do i have to use the quotient rule on this part

17. Mar 6, 2007

### dontdisturbmycircles

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 ya go... Do you understand now?

Last edited: Mar 6, 2007
18. Mar 6, 2007

### afcwestwarrior

ok can u demonstrate it too me

19. Mar 6, 2007

### afcwestwarrior

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)

20. Mar 6, 2007

### dontdisturbmycircles

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: Mar 6, 2007