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Homework Help: Equation of tangent - Implicit or Partial DifferentiatioN?

  1. Jun 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Need to find the tangent to the curve at: e^(xy) + x^2*y - (y-x)^2 + 3

    I just implicitly differentiate the expression to find the gradient and then use the points given to find the equation, right?
    Or does this involve partial differentiation?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 11, 2012 #2
    None of those :wink:

    Just differentiate with respect to x, using the chain rule and product rule.
  4. Jun 11, 2012 #3


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    Is it a curve e^(xy) + x^2*y - (y-x)^2 + 3=constant, or a surface F(x,y)=e^(xy) + x^2*y - (y-x)^2 + 3? And at which point do you need to find the tangent or gradient?

  5. Jun 11, 2012 #4
    curve at (0,2)
    how do you differentiate with respects to "x" if there's a "y" without one of those methods..
  6. Jun 11, 2012 #5
    As I said, use product and chain rules. For example, the derivative of the first term would be,

    [tex]\frac{de^{xy}}{dx} = e^{xy} (y + x\cdot \frac{dy}{dx})[/tex]
  7. Jun 11, 2012 #6
    ah I see
    would it have been wrong to implicit differentiation then?
  8. Jun 11, 2012 #7
    Implicit differentiation is used when you cannot differentiate the terms using these methods, specifically when you have to integrate function raised to another function of the same variable. This is not the case with the given curve, so there is no need for implicit differentiation.
  9. Jun 12, 2012 #8


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    Infinitum, what you did, it is implicit differentiation.

    y is given as a function of x implicitly by an equation R(x, y) = 0. We differentiate R(x, y) with respect to x and then with respect to y, multiplying it with dy/dx.

  10. Jun 12, 2012 #9


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    What, exactly, do you mean by "implicit differentiation"? It is clearly called for in this problem.
  11. Jun 13, 2012 #10
    You're right. I don't know, but absent-mindedly, I interchange definitions sometimes....:surprised

    Apologies, dan38!
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