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Derivative Implicitly Date:12/1/07 Long Zhao Teacher:Ms.Delacruz In calculus, a method called implicit differentiation can be applied to implicitly defined functions. This method is an application of the chain rule allowing one to calculate the derivative of a function given implicitly. As explained in the introduction, y can be given as a function of x implicitly rather than explicitly. When we have an equation R(x,y) = 0, we may be able to solve it for y and then differentiate. However, sometimes it is simpler to differentiate R(x,y) with respect to x and then solve for dy / dx. Source:Wikipedia dy Find of dx x²+5x+3y²=21 Step 1:Take derivative on both sides dy (x²)+dy(3y²)= dy (21) dx dx dx Step 2:Get the derivative 2x+6y dy =0 dx dy Step 3:Solve for dy =-2x/6y dx dx To find the equation of the tangent line of We find the derivative The equation of the tangent line is: y - 1 = 0.25 (x - 2) Source:http://archives.math.utk.edu/v isual.calculus/3/implicit.5/index.html An example of an implicit function, for which implicit differentiation might be easier than attempting to use explicit differentiation, is x³+2y²=8 In order to differentiate this explicitly, one would have to obtain (via algebra) Y=± 8 x 2 4 and then differentiate this function. This creates two derivatives: one for y > 0 and another for y < 0. One might find it substantially easier to implicitly differentiate the implicit function; One might find it substantially easier to implicitly differentiate the implicit dy function;4x³+4y =0 dx thus, dy 4 x 3 x 3 dx 4y y Source:wikipedia