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Derivatives
- Thread starter kggirl
- Start date
Find dy/dx if xy^2 + x^2y + y = 0
Is this correct:
x2y + 2xy +y
Sorry, this is not correct. You'll need to use implicit differentiation. On the first term, for example:
The derivative of xy^2 is x*2y*(dy/dx) + (y^2)*1. Can you handle the rest?
wjm11 said:Find dy/dx if xy^2 + x^2y + y = 0
Is this correct:
x2y + 2xy +y
Sorry, this is not correct. You'll need to use implicit differentiation. On the first term, for example:
The derivative of xy^2 is x*2y*(dy/dx) + (y^2)*1. Can you handle the rest?
I'm not sure I get this, so would it be?
x(2y) + 2(x) y+y =0
x2y + 2xy +y^2 = 0
I'm not sure I get this, so would it be?
x(2y) + 2(x) y+y =0
x2y + 2xy +y^2 = 0
Still not correct. You need to read your book.
Example: The derivative of y^2 with respect to x is 2y*(dy/dx).
In your problem you need to differentiate each of the three terms, separate out the terms that have (dy/dx) in them, and finally, solve in terms of dy/dx.
wjm11 said:I'm not sure I get this, so would it be?
x(2y) + 2(x) y+y =0
x2y + 2xy +y^2 = 0
Still not correct. You need to read your book.
Example: The derivative of y^2 with respect to x is 2y*(dy/dx).
In your problem you need to differentiate each of the three terms, separate out the terms that have (dy/dx) in them, and finally, solve in terms of dy/dx.
x2y + 2xy + 0 =
2x + 2xy + 2x + 2xy = 0
4x + 2xy =0
What is the derivative of 2xy with respect to x?
We have a product: (2x) * (y)
So we must use the product rule:
If this isn't ringing any bells then you need to consult with your textbook/notes from class on implicit differentiation.
We have a product: (2x) * (y)
So we must use the product rule:
Code:
d d d
---- 2xy = ---- (2x) * y + ---(y) * 2x
dx dx dx
dy
= 2 * y + ---- * 2x
dx
dy
= 2y + 2x ---
dx