Implicit Differentiation

~A very important technique in Calculus

~It enables us to analyize equations which are not functions of x where y or f(x) is not explictly expressed

~Related rate problems depend upon this technique as well

~It also enables us to derive most derivative formulas for the transcendental functions (i.e., trig, inverse trig, exponentials, & logs)

~Consequently, with this technique, our derivative formula base is expanded greatly

~To understand this technique for getting derivatives, we need to understand the essence of the derivative operator, d( )/dx. This operator takes the derivative of an expression with respect to x. The operator d( )/dt will do the same but with respect to t. Whatever variable is in the operator (bottom), that's what you are differentiating with repect to...you must keep this in mind always.

~When the x derivative operator meets an expression that contains y, a dy/dx will be produced. Since dy/dx is what you want, forgetting this, destroys your entire problem.

~Let's practice a little

d(y)/dx = dy/dx, d(x)/dx = 1, d(y²)/dx = 2y d(y)/dx = 2ydy/dx. (we are using the power rule here)

d(x³)/dx = 3x² (no dy/dx pops out ), d(y)/dt = dy/dt, d(x)/dt = dx/dt, d(t)/dt = 1, d(x²)/dt = 2xdx/dt

d(x+y)/dt = dx/dt + dy/dt, d(A)/dt = dA/dt, d(xy)/dx = xdy/dx + y (1) or simply xdy/dx +y (product rule is used here)

~Let's use the time derivative operator and operate on both sides with the equation for the Area of a triangle with base b & height h

d(A)/dt = d[(1/2)bh]/dt, this gives, dA/dt = (1/2)[bdh/dt + h db/dt] (the product rule was used on bh, assuming they are not fixed constants)

~Let's operate on the same equation using the x operator, d(A)/dx = d[(1/2)bh]/dx, this gives, dA/dx = (1/2)[bdh/dx + h db/dx] (a similar result)

~Most of the time we are interested in getting dy/dx, so the x operator is used. In most cases, since the equation is not solved for y, dy/dx will not be expressed explictly. So, after we do our operation, we would have to use a little alegbra to isolate dy/dx

~Here is an exmple of what I mean:

~Find dy/dx for xsin(y) = y²cos(x)

~Since we are after dy/dx, we will operate using d/dx, the x operator

~Remember, when this operator meets any expression that contains a y in any form, a dy/dx will be produced

~Let's operate by taking the derviatve of both sides of the equation

~We get, d(xsin(y))/dx = d(y²cos(x))/dx

continuing, x d(sin(y))/dx + sin(y)d(x)/dx = y²d(cos(x))/dx + cos(x)d(y²)/dx (product rule on both sides)

we now get, xcos(y)dy/dx + sin(y) = y²(-sin(x)) + cos(x)(2ydy/dx) (remember when you expect dy/dx to appear)

~Now, all that is left is to solve for dy/dx (this requires good algebra skills)

~First, bring the terms that contain dy/dx on the same side of the equation and the ones that do not on the other side

Let's bring all dy/dx terms to the right side & the others to the left side

~This is what results: sin(y) + y²sin(x) = cos(x)(2ydy/dx) - xcos(y)dy/dx

~Now, factor dy/dx out of the two terms on the right side, sin(y) + y²sin(x) = dy/dx[2ycos(x)-xcos(y)]

~Now, divide by the factor next to dy/dx to isolate dy/dx on the right side

~That results in the completion of our problem

~So, [sin(y) + y²sin(x)]/[2ycos(x)-xcos(y)] = dy/dx

~Like most everything else in calculus, to gain proficiency, one needs to spend time & effort practicing these techniques