Task: Find an equation for the line tangent to the graph of x³ + 2xy² + y³ = 1 at the point (1, -2).
 

The problem is that y is not explicitly a function of x. It turns out that it is possible to find some function whose graph coincides with the graph of
x³ + 2xy² + y³ = 1 near the point (1, -2). We say that the given formula defines yimplicitly as a function of x.
 

There are two obvious approaches to finding the tangent line: (1) You could solve for y and use traditional methods; or (2) you could graph the curve and use the calculator to find the tangent line. Unfortunately, neither method works. (1) fails because the calculator won't solve the equation. (2) fails because, although the calculator can graph implicitly defined functions (though not very well), the automatic tangent line doesn't work in that mode.

So we need a non-obvious approach. Here are three.
 

Linearization approach

• First, enter the equation and store it in some convenient variable, such as e for equation. (Be careful! Type x^3+2x*y^2+y^3=1; if you leave out the *, the calculator thinks the middle term is the square of the two letter variable xy.)
• We're assuming that y is a function of x and that we can find a tangent line to the graph of the equation at (1, -2). Any such line has an equation of the form y = m*(x-1) - 2, where m is the slope. The tangent line is supposed to be a good approximation to the graph of function, so let's assume the tangent line is the graph of the function.
• Store the expression m*(x-1)-2 in the variable y, and now see what equation e looks like.
• The quickest way to find the value of m when x=1 is to differentiate both sides of the resulting equation. d (e,x)|x=1 will do the trick. It should be routine to solve the resulting equation for m, but the calculator will do the solving for you if you like.

 

• The textbook's method is exactly the same as the one described above, but what you write down is a little different. Instead of substituting the formula for a general line through (1, -2) for y, we just assume y has that form, and differentiate. Since we don't know the derivative of y, we just write down y'  when we need it.
• Differentiate both sides of the equation x³ + 2xy² + y³ = 1 with respect to x, assuming y is a function of x. Don't forget to use the product rule and the chain rule: the derivative of y³ is 3y² * y' and the derivative of 2xy² is 2y² + 4xy * y' .
• Then substitute 1 for x and -2 for y and solve for y' .

 

• There's no implicit differentiation function built into the calculator, but it's easy to write one. Unfortunately, if you want to understand why the formula below is correct, you'll probably have to read chapter 11 of the text.
• Define impdif(z,x,y) = d (left(z)-right(z),x)/d (right(z)-left(z),y). z is the equation (It must be an equation!), x is the input variable, and y is the output variable.