How do you implicitly differentiate #xy^2-x/y=-1#?
Also the first term will need the use of the product rule.
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To implicitly differentiate the equation xy^2 - x/y = -1 with respect to x, follow these steps:
- Differentiate each term of the equation with respect to x.
- Apply the product rule for the term xy^2 and the quotient rule for the term x/y.
- Solve for dy/dx, which represents the derivative of y with respect to x.
Here are the steps:
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Differentiate each term: For xy^2, apply the product rule: (uv)' = u'v + uv' For x/y, apply the quotient rule: (u/v)' = (u'v - uv') / v^2
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Applying the product rule to xy^2: (xy^2)' = x(dy/dx)(2y) + y^2 = 2xy(dy/dx) + y^2
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Applying the quotient rule to x/y: (x/y)' = ((1)(y) - (x)(1/y^2))(1/y^2) = (y + x/y^2)(1/y^2) = (y^3 + x) / y^3
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Re-write the original equation: xy^2 - x/y = -1
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Now, differentiate each term with respect to x: (2xy(dy/dx) + y^2) - ((y^3 + x) / y^3) = 0
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Simplify and solve for dy/dx: 2xy(dy/dx) + y^2 - (y^3 + x) / y^3 = 0 2xy(dy/dx) + y^2 - (y^3 + x) / y^3 = 0 2xy(dy/dx) + y^2y^3 - (y^3 + x) / y^3 = 0 2xy(dy/dx) + y^5 - (y^3 + x) / y^3 = 0 2xy(dy/dx) + y^5 - (y^3 + x) / y^3 = 0 2xy(dy/dx) + y^5 - (y^3 + x) / y^3 = 0 2xy(dy/dx) + y^5 - y^5 - x/y^3 = 0 2xy(dy/dx) - x/y^3 = 0 2xy(dy/dx) = x/y^3 dy/dx = x / (2y^4)
Therefore, the implicit derivative of xy^2 - x/y = -1 with respect to x is dy/dx = x / (2y^4).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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