© 2002 by Karl Hahn

x^{2}y + y^{3} + x = 11 eq. 5.61One of the things they might ask you, for example, is what is the slope of this curve at the point, (1,2)? Recall that implicit differentiation does not give you the derivative of something outright (that is, as a function of x), but it does tell you something about the derivative of a curve like this. So lets take the implicit derivative of this equation.
The first term is x^{2}y. Clearly you have to use the product rule to take its derivative. The derivative of x^{2} is 2x. The derivative of y is y'. So applying the product rule to the first term only, you get 2xy + x^{2}y'.
The second term is y^{3}. Using what you learned about implicit differentiation, you can see that the derivative of this term is 3y^{2}y'.
The third term is x, and you already know that the derivative of x is 1. And the last term (the one to the right of the equal) is a constant, so its derivative is zero. Putting it all together, when you take the derivative of equation 5.61, you get
2xy + x^{2}y' + 3y^{2}y' + 1 = 0 eq. 5.62aSo how do you use this to answer the question of what the slope of the curve is at (1,2)? First, solve for y' in terms of x and y. Do this using a little algebra. Move all the terms of the derivative equation (eq. 5.62a) that have a y' factor over to one side of the equation.
x^{2}y' + 3y^{2}y' = 2xy  1 eq. 5.62bNow factor the y' out of that side of the equation:
y'(x^{2} + 3y^{2}) = 2xy  1 eq. 5.62cand then divide both sides by the other factor:
2xy  1 y' =The point in question was (1,2). So in equation 5.62d, substitute 1 for x and 2 for y:eq. 5.62d x^{2} + 3y^{2}
4  1 5 at (1,2): y' =Once you have the implicit derivative and you have isolated y', this kind of problem is pretty easy. It's just plugging in the numbers.= eq. 5.63 1 + 12 13
In the figure, you can see the curve along with the tangent line
at the point, (1,2). See if you can figure out how I determined
the equation of that line.
The other kind of problem they like to give you is to ask you where the horizontal and vertical tangents are to the curve. We will do this with
x^{2} + 5xy + 20y^{2} = 100 eq. 5.64Of course, we will need to do the implicit differentiation:
2x + 5y + 5xy' + 40yy' = 0 eq. 5.65First we find all the horizontal tangents. How do we do that? When the tangent line is horizontal, its slope is zero. That means that y' at the point of tangency must also be zero. So we set y' = 0, and see what happens to our derivative equation. Notice that all the terms that have a y' as a factor drop out and you are left with:
2x + 5y = 0 eq. 5.66aor equivalently
2 y = Now substitute this expression for y back into the original equation (eq 5.65) and solve for x.x eq. 5.66b 5
16 x^{2}  2x^{2} +Simplifying you getx^{2} = 100 eq. 5.67a 5
11 500Take the square root to find x.x^{2} = 100 or x^{2} =eq. 5.67b 5 11
___ √500 x = ±That gives you the x coordinates for two points of tangency. Put those x's back into equation 5.66b to find the corresponding y's. That's iteq. 5.67c √11
So now, how do you find the vertical tangents? The slope of a vertical line doesn't exist because when you try to find it, you get a zero in the denominator. As you draw lines that are closer and closer to the vertical (with positive slope), the slope goes to infinity. This offers a method for finding vertical points of tangency. We take the implicit derivative equation, and take the limit as y' goes to infinity.
lim 2x + 5y + 5xy' + 40yy' = lim 0 eq. 5.68a y' → ∞What happens here is that as y' gets huge, all the terms that have no y' factor in them become insignificant, including those to the right of the equal. Another way of looking at it is if you divide the whole thing through by y':
lim y' → ∞ 
2x 5y 
5xy' + 40yy' = 0 eq. 5.69aAnd now divide out the y', and you're left with
5x + 40y = 0 eq. 5.69b 5 y = Now simply substitute this expression for y back into equation 5.64, and solve for x.x eq. 5.69c 40
25 500 x^{2} That gives you the x coordinates of the two vertical points of tangency. To get the corresponding y values, just take these x's, and plug them back into equation 5.69c.x^{2} +x^{2} = 100 eq. 5.610a 40 1600 11 1600x^{2} = 100 or x^{2} =eq. 5.610b 16 11 40 x = ±__eq. 5.610c √11
Now suppose they ask you to find the tangent lines whose slopes are 1 (or any other value for that matter). Simply go back to equation 5.65, and plug in 1 (or whatever other slope value they are asking for) for y'.
2x + 5y + 5x + 40y = 7x + 45y = 0 eq. 5.611a 7 y = I'll leave it to you to finish this part of the problem in the same way as I did the other two parts  that is, substitute this expression for y back into equation 5.64, and solve for x.x eq. 5.611b 45
You can see the graph of all this to the right. The blue labels a vertical tangency point.
The brown labels a horizontal tangency point. You get to fill in a black
tangency point (one with slope of 1) yourself.
Move on to Exponentials and Logs