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Math Help - differentiation

  1. #1
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    differentiation

    Find the equation using the tangent line to the curve defined by the equation at the specific point.

    (x^2+y^2)^2= 4x^2y; (-1,1)


    Use implicit differentiation to verify that the equation implicitly defines a solution of the differential equation.

    y^2= 1+Ce^-2x; y(y'+y)=1

    My teacher just assigned theses and did not explain what to do im lost please help. I believe derivatives are used as well as the chain rule...i think
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  2. #2
    Member Haven's Avatar
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    Implicit differentiation can be very tricky the first time through.

    I'll try to explain it the best that I can with your first example

    So we have:
    (x^2+y^2)^2= 4x^{2y}
    Simplify it a little bit:
    (x^2+y^2)= 4x^y
    Now we take derivatives of both sides
    \frac{d}{dx}(x^2+y^2)= \frac{d}{dx}(4x^y)
    Now here's the tricky part. Any time we take the derivative of a multivariable function like this one we consider y to be a function of x. So for the sake of understanding how implicit differentiation works, I'll replace every occurence of y with y(x). However, this isn't what you should be doing in your homework.
    \frac{d}{dx}(x^2+y(x)^2)= \frac{d}{dx}(4x^{y(x)})
    Now we use properties of the derivative to make our job easier:
    \frac{d}{dx}(x^2)+\frac{d}{dx}(y(x))^2= \frac{d}{dx}(4x^{y(x)})
    Notice the first term contains only x, so we can take our usual derivative:
    2x+\frac{d}{dx}(y(x))^2= \frac{d}{dx}(4x^{y(x)})
    However, the other terms contain y(x), so we have to use chain rule:
    Remember: \frac{d}{dx}(y(x)) = \frac{dy(x)}{dx} * \frac{d}{dy(x)}
    So whenever we see a y(x) (or rather a y), we take the derivative with respect to why and multiply it by \frac{dy(x)}{dx}
    So using the above argument and chain rule
    2x+\frac{dy(x)}{dx}*(y(x)^2)'= 4(\frac{dy(x)}{dx}*y'(x)*x^{y(x)}\log{x} + \frac{x^{y(x)}}{x})
    Now we replace y(x) with y and take derivatives:
    2x+\frac{dy}{dx}*2y= 4*(\frac{dy}{dx}*1*x^{y}\log{x} + \frac{x^{y}}{x})
    And now you just solve algebraically for \frac{dy}{dx}
    Last edited by Haven; November 19th 2009 at 10:16 PM.
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