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Homework Help: Implicit Derivatives - Trigonometry

  1. Apr 14, 2008 #1
    Hello hello everyone...
    I have a problem that I've been trying to figure out for a couple of days now. Its a question off of an assignment, and I have been busting my balls(excuse my french) with no results. HELP would be GREATLY appreciated...lthis is the question at hand:

    xsin(y) + x^3 = tan^(-1)(y)
    the second part is an inverse

    I have to find the first and second derivative, critical values, points of inflection, intercept, in order to sketch it afterwards.

    I shall cross my fingers with the hope that I will find genius out there :D !!!
  2. jcsd
  3. Apr 15, 2008 #2
    use leibniz notation d/dx everything.
    if you d(x^3)/dx , u get 3x^2.
    The implicit part is differentiating the ys.If you assumme sin (y) = u --> du/dx = d(sin(y))/dx , is that clear?

    now du/dx = du/dy . dy/dx becuase u is composed of the variable y. so if you do this you'll have du/dx = d(sin(y))/y . dy/dx. Leave dy/dx and d(sin(y))/y is cos(y).
    so d(sin(y))/dx = cos(y) . dy/dx.

    The rest is just applications of this, with product rule and everything. arctan (the inverse tan) has a derivative of (1/(1+x^2)). If you understand the chain rule it should be easy. Once you take the first derivative of everything, you can isolate the equation for dy/dx to be on one side and everything on the other to find critical values.

    If you have any specific questions feel free to ask, good luck.
  4. Apr 16, 2008 #3
    thanks for the reply it makes a lot of sense
    The last question I have now is... how would I go about finding the x-y intercepts, domain, range, asymptotes and symmetry of the function??
    By getting the derivative, I can find the critical values, local maximums/minimums, inflection points, and concavity intervals... but I also need to sketch the curve and in order to do so I would need the preceding info as well. I dont know how to isolate the y on one side, and the x on the other in order to do so :(


    ps. Would I find the second derivative applying the same procedure (implicit differentiation)??

    Your help is greatly GREATLY appreciated!!!!!!! :D
  5. Apr 18, 2008 #4
    The second derivative can be found the same process, just d(whatever is on each other side) by dx.

    Let's say you have x^3 + y^3 = xy

    First derivative is : 3x^2 + 3y^2 . dy/dx = y + x . dy/dx
    Now if everything on left side is g(x) and the right side is n(x). Then just do dg(x) / dx and do dn(x) / dx.

    So it would be : d[3x^2 + 3y^2 . dy/dx] / dx = d[y + x . dy/dx] / dx

    6x + 3y^2 . (d^2y / dx^2) + 6y . (dy/dx) ^ 2 = 2 . dy/dx + x . (d^2y / dx^2)

    That is the second derivative. Notice (dy/dx) ^ 2 is not same as (d^2y / dx^2).

    One means the square of first derivative, second is notation for second derivative.

    Domain is a concept, has no equations. Ok the first term has a sin, and all real numbers are ok in sin functions so you're good. The second term , x^3 not a problem, all real numbers. The right hand side has a tan^-1 which means it's domain would be from one asymptote of tan(x) to another. So y has to be from -pi/2 to pi/2 and that is the domain of y values in the function.

    X-int means when y equals to 0, what can x be. Just replace all ys with zero.

    Y-int is opposite, replace all xs with zero, solve for y.

    I don't think there is any asymptotes in the function tho, there is no denominators.
    Last edited: Apr 18, 2008
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