Implicit Differentiation

J

JoeJ

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Jun 20, 2005
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I'm having problems on these equations, and I was wondering if I could get any help...

Use implicit differentiation to find dy/dx.

1. x^3 - xy + y^2 = 4

2. (x^2)y + (y^2)x = -2

3. (sin (pi)x + cos (pi)y)^2 = 2

4. sin x = x(1 + tan y)

5. y = sin(xy)

Please show step by step how each is worked out.

Thank you.
 
JoeJ said:
Use implicit differentiation to find dy/dx.

1. x^3 - xy + y^2 = 4
Click to expand...
Think Chain Rule. Chain Rule. Say it out loud 12 times. Chain Rule.

We don't know what y' is, so when it comes to finding the derivative of y, we'll just write y'.

One term at a time.

(d/dx)(x^3) = 3*x^2 -- OK. That one was easy.

(d/dx)(x*y) = x*y' + y*1 = x*y' + y -- Needed the product rule, too.

(d/dx)(y^2) = 2*y*y' -- Were you thinking about the chain rule?

(d/dx)(4) = 0

3*x^2 - (x*y' + y) + 2*y*y' = 0

Solve for y'

3*x^2 - x*y' - y + 2*y*y' = 0

You do the rest.

Did I mention the Chain Rule?
 
Ok, for that one, I got y' = (-3 x^2 - y) / (x + 2y). I was wondering if that's correct?
 
Why wonder? Work slowly and deliberately and gain confidence.

3*x^2 - x*y' - y + 2*y*y' = 0
- x*y' + 2*y*y' = -3*x^2 + y
(- x + 2*y)*y' = -3*x^2 + y

Do you get a different answer, now?
 
That's what I thought. Now, for the next one, x^2 *y + y^2*x = -2, here's what I figured out...

(dy/dx) x^2 *y + (dy/dx) y^2 *x = (dy/dx)(2)
2xyy' + 2yy'x = 0
y'(2xy + 2yx) = 0
y' = (0/(2xy + 2yx))
y' = 0
 
Did we settle the first one? You have all negative in the numerator and all positive in the denominator. I seem to have one of each in both.

Moving on, I guess...

2. (x^2)y + (y^2)x = -2

Product Rule ring any bells?

(d/dx)(x^2*y) = x^2*y' + y*(2*x)

(d/dx)(y^2*x) = y^2*1 + x*(2*y)*y'

x^2*y' + y*(2*x) + y^2 + x*(2*y)*y' = 0

I have four terms and you only two. You'd better take another look at it, slowly and deliberately.
 
For the first one, do have negatives in both...my answer is y' = ((-3x ^ 2) - y) / (-x + 2y)...I do have a negative in both as well. But, you said you had one in the numerator? For some reason, I got two.

For the second, I got...
y'(x^2 + x(2y)) = -y(2x) + y2
y' = (-y(2x)+ y^2)/(x^2 + x(2y)))
 
You have now made the exact same error twice. I think I get it.

Given this situation:

-A+B

You seem to be factoring out a factor of -1. It SHOULD appear as this:

-(A-B)

Unfortunately, you keep writing it without the parentheses so it looks like you are not doing anything except putting the wrong sign on the second term.

From here:

x^2*y' + y*(2*x) + y^2 + x*(2*y)*y' = 0

Get

x^2*y' + x*(2*y)*y' = -2*x*y - y^2

OR

x^2*y' + x*(2*y)*y' = -(2*x*y + y^2) <== Without the parentheses, it is just wrong.

From here:

3*x^2 - x*y' - y + 2*y*y' = 0

Get

-x*y' + 2*y*y' = -3*x^2 + y

OR

-x*y' + 2*y*y' = -(3*x^2 - y) <== Without the parentheses, it is just wrong.

OK, let's do the next one. Your turn.
 
Ok, here's my steps I get for the second one...

x ^ 2 * y + y ^ 2 * x = -2
(d / dx)x ^ 2 * y + (d / dx) y ^ 2 x = (d / dx)(2)
(d / dx)y ^ 2 * x = y ^ 2 * 1 + x (2*y) * y'
x ^ 2 * y' + y (2 x) + y ^ 2 + x (2*y) y' = 0
y '(x ^ 2 + x (2*y)) = -y (2*x) + y ^ 2
y' = (-y * (2x) + y ^ 2) / (x ^ 2 + x (2*y))
 
JoeJ said:
x ^ 2 * y' + y (2 x) + y ^ 2 + x (2*y) y' = 0
y '(x ^ 2 + x (2*y)) = -y (2*x) + y ^ 2
Click to expand...
You did it again! Did you read my last post? How did you get +y^2 on both sides?

The sign changes on the 2*x*y part. Why doesn't it change on the y^2 part?
 
Ok, I don't know why I keep missing something simple like that. I
keep on just looking over, and copying down from the left side, without realizing what I'm doing, I guess. It should be then y ' = (-y * (2x) - y ^ 2) / (x ^ 2 + x * (2 y))
 
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