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.

- Thread starter JoeJ
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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.

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Think Chain Rule. Chain Rule. Say it out loud 12 times. Chain Rule.JoeJ said:Use implicit differentiation to find dy/dx.

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

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?

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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 second, I got...

y'(x^2 + x(2y)) = -y(2x) + y2

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

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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.

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))

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You did it again! Did you read my last post? How did you getJoeJ 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

The sign changes on the 2*x*y part. Why doesn't it change on the y^2 part?