Implicit differentiation What to do when x = f(y) (or worse)

A expression is often written explicitly, in the familiar form y = f(x). Sometimes this is not easily done; sometimes it is just not possible (because the expression in question might not even be a function). But the need to differentiate is still strong within us.

Implicitly? A relation F(x,y) = 0 is said to define the function y = f(x) implicitly if, for x in the domain of f, F(x, y) = 0. To find f’(x) in terms of x and y, we must unleash the True Power of the Chain Rule.

You have already observed this: when you found the derivatives of the inverse trig functions. But did you notice this: If x = sin y then dy/dx = 1/cos y And if y = sin x, then dy/dx = cos x ???

Implicit differentiation Similarly for cosine: If x = cos y then dy/dx = - 1/sin y And if y = cos x, then dy/dx = - sin x ???

Sounds like a theorem: One could speculate that this is not a coincidence and that the pattern which generalizes might even be worthy of a name. Hint: Its called the Inverse Function Theorem: if f and g are differentiable inverse functions at any x in the domain of g, g’(x) = 1/ f’(g(x)), where f’(g(x)) ¹ 0.

So much for theory. How would you find the slope of the curve: x2 – 2y3 + 4y = 2, at the point (0,1)? You could try to solve for y: y(4 - 2y2) = 2 – x2 or y(2 - yÖ2)(2 + yÖ2) = 2 - x2 Yuck!

Or you could solve for x: Slope of x2 – 2y3 + 4y = 2, at (0,1)? OR: Solve for x: x2 = 2y3 - 4y + 2 x = +/- [y3 - 4y + 2]1/2 Graph (x, y) and find ‘inverse slope’: Looks like 1/slope = 0 at x = 0?

Another way: Plot both x and y in terms of the “dummy variable” t Slope of x2 – 2y3 + 4y = 2, at (0,1) x = +/- [t3 – 4t + 2]1/2 y = t Mathematica calls this a ParametricPlot

A better way: Differentiate both sides “implicitly” Slope of x2 – 2y3 + 4y = 2, at (0,1)? Take dy/dx of both sides of the equation: 2x – 6y2 dy/dx + 4 dy/dx = 0 Solve for dy/dx : 2x = (6y2 - 4) dy/dx dy/dx = 2x / (6y2 - 4) At (0,1), dy/dx = 0. Voila!

Explicit rules on how to be implicit Be sure to differentiate both sides of the equation with respect to (wrt) x. Remember all applicable rules (product, quotient, chain). Variables ‘agree’: d/dx [x3] = 3 x2. Variables ‘disagree’: use your rules. d/dx [y3] = 3 y2 dy/dx by chain rule.

Explicit rules on how to be implicit Collect the terms involving dy/dx , so they are on the same side of the equation. Treat dy/dx like the ‘variable’ for which you are solving.

Example: y3 + y2 - 5y - x2 = -4 3y2 dy/dx + 2y dy/dx –5 dy/dx – 2x = 0 dy/dx (3y2 + 2y – 5) = 2x dy/dx = 2x / (3y2 + 2y – 5) Do not be alarmed by the presence of both x and y in your equation for dy/dx .

And now for the 2nd derivative: dy/dx = 2x / (3y2 + 2y – 5) d2y = [2(3y2 + 2y – 5) – 2x(6y +2) dy/dx] dx2 (3y2 + 2y – 5)2 which you can simplify, if you like.

Circle: x2 + y2 = r2 dy/dx = -2 x / (2 y) = - x / y 2x + 2y dy/dx = 0 This is the slope of the tangent to any circle, defined at any point where y ¹ 0.

Circle: x2 + y2 = r2 x<0, y >0, slope > 0 Notice how the slope changes signs from quadrant to quadrant and that dy/dx is independent of r. x<0, y <0, slope < 0

Circle: x2 + y2 = r2 dy/dx = -2 x / (2 y) = - x / y Can you show that the 2nd derivative of a circle of radius r = - r2/y3 ? This expression allows the concavity of a circle to change sign: Think of the top half of the circle as concave down; the bottom half as concave up.

Ellipse: 3x2 + 4y2 = 9 dy/dx = -3 x / 4 y 6x + 8y dy/dx = 0 Once again, only defined where y ¹ 0. Notice the coefficient and its relationship to the eccentricity of the ellipse.

Parabola: x + y2 = 1 dy/dx = -1 / 2 y 1+ 2y dy/dx = 0 Once again, only defined where y ¹ 0.

Hyperbola: xy = 4 Which almost leads us to…. dy/dx = - y / x y + x dy/dx = 0 dy/dx = - y / x Compare this to the derivative of the circle, dy/dx = - x / y. Which almost leads us to….

Orthogonal trajectories Two curves are orthogonal if, at their point of intersection, their tangent lines are mutually perpendicular.

Orthogonal trajectories To show orthogonality, first find the intersection points of the two curves. Then determine if their derivatives are negative reciprocals at the intersection points.

Example: 2x2 + y2 = 6 and y2 = 4x dy/dx = -2x/y. At (1,2), dy/dx = -1. Intersection: 2x2 + 4x = 6 or x2 + 2x – 3 = 0, or (x + 3)(x - 1) = 0 Roots: x = -3, where y2 = 4(-3) is imaginary. x = 1, where y2 = 4x = 4. So the only real intersection is (1, 2). Slopes: 4x + 2y dy/dx = 0 dy/dx = -2x/y. At (1,2), dy/dx = -1. 2y dy/dx = 4 dy/dx = 4/2y. At (1,2), dy/dx = 1. Voila!

Now for some practice!