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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    i know how to get the derivative of this one, but i don't know what to do from there:

    f(x) + x^2[f(x)]^3 = 10 and f(1) = 2, find f^(prime) (1)

    i also don't know how to differentiate these, and find an equation of the tangent line to the curve at the given point:

    1. x^2 + y^2 = (2x^2 + 2y^2 -x)^2 at (0, 1/2)

    2. x^(2/3) + y^(2/3) = 4 at (-3sqrt3, 1)

    3. 2(x^2 to y^2)^2 = 25(x^2 - y^2) at (3,1)

    4. y^2(y^2 -4) = x^2(x^2 -5) at (0, -2)

    if anyone could help me, i would be soo grateful...thanks!!
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  2. #2
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    Feb 2008
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    Westwood, Los Angeles, CA
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    Thanks
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    1st question response

    For your first question, consider it written as:

    y + (x^2)(y^3) = 10.

    For the derivative, I got y' = -2x(y^3)/((3(x^2)(y^2) + 1))

    You want to find f'(1), so you'll need to substitute x=1 into your derivative. Since you did implicit differentiation, though, you also need to have a value to substitute in for y. For that, use the fact that your original function passes through (1,2) (since f(1) = 2). You're finding the derivative of your function at that point, so substitute x = 1 and y = 2 into your derivative. That's it.

    For your second question, the one you labeled "1.", find the derivative just like you did in the first one, implicitly.

    I get:

    2x + 2yy' = 2(2x^2 + 2y^2 - x)(4x + 4yy' -1) [using the chain rule on the right]

    Now, plug in your x and y values (0 and 1/2, resp.) and then solve for y'. Remember that your derivative, y', gives you the slope of the tangent line to your function at a particular point, in this case (0, 1/2). So now you know the slope of the tangent line (f'(0), the y' from above) and you also know a point on the tangent line (0, 1/2). Since you know a point and the slope, use point-slope form to write the equation of the line.

    Point-slope form:

    (y - y_1) = m (x - x_1).

    Use the coordinates of your point for (x_1, y_1) and the slope for m.

    Hope this helps - I'll leave the last few for you. Once you find the derivatives, the idea behind finding the equation of the tangent line is the same - use point-slope form.
    Last edited by Mathnasium; February 26th 2008 at 10:08 PM.
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