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Calculus/AP Calculus Implicit Differentiation, Vertical Tangents, and Second Derivatives


This is a two-part question:

First, how do you find the derivative of y+cos(y)=x+1? I know you need to use implicit differentiation, but I don't understand how you would use it in this case.

Second, how would you determine the coordinate points for the derivative's vertical tangents?

For y+cos y =x+1, when you differentiate both sides with respect to x, you will have

   (1-siny) dy/dx = 1 =====>  dy/dx = 1/ (1-siny)

Note that implicit differentiation is carried out on the LHS of the above because
the expression for the LHS is purely given in terms of y, and you are differentiating indirectly wrt x.

If this still bothers you, here is a more fundamental explanation for how I obtained the LHS of the above:

   d/dx ( y+ cos y) = d/dy (y+cosy) * (dy/dx) = (1-siny) dy/dx

When the derivative has vertical tangents, it simply means the gradient function dy/dx equals infinity, or dx/dy =0

In this case, we set 1-siny =0 , ie siny =1  which therefore means y=pi/2 , x =pi/2 -1 (this is obtained through substituting the y value found into the original expression)

Hope this helps. Peace.


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Frederick Koh


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