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Implicit Differentiation

Rules for Implicit Differentiation

  1. Differentiate both sides with respect to x
  2. Collected all terms involving dy/dx or y' and move all of the other terms to the right side
  3. Factor dy/dx out of the left side, and solve for it

Example problem

Find the tangent line to the circle x2 + y2 = 1 at the point (√2/2, √2/2).

You could solve for y, and find the answer that way, or you could implicitly differentiate.

x2 + y2 = 1

dy/dx(x2 + y2) = dy/dx(1)

2x + 2yy' = 0 (note that the chain rule was needed)

2yy' = -2x, y' = -x/y

So, at (√2/2, √2/2), y' = -1

Logarithmic Differentiation

This is a type of implicit differentiation, where you take the natural log of both sides before implicitly differentiating.

Initially, this looks really difficult, you'd have to do the product rule, then the quotient rule, and that would be very annoying. Luckily, by doing logarithmic differentiation, this isn't that bad.

ln(y) = ln(3x-2) + ln(3x2 + 4) + ln(2x5 - 8) - 1/2ln(x - 1)

Now, you can differentiate both sides:

y'/y = 3/(3x-2) + 6x/(3x2 + 4) + 10x4/(2x5 - 8) - 1/(2x - 2)

y' = y(3/(3x-2) + 6x/(3x2 + 4) + 10x4/(2x5 - 8) - 1/(2x - 2))

y' = (3x-2)(3x2 + 4)(2x5 - 8)/√(x-1) * (3/(3x-2) + 6x/(3x2 + 4) + 10x4/(2x5 - 8) - 1/(2x - 2))

So, although this becomes extremely ugly, it wasn't very difficult. It's just important to remember that dy/dx(ln(y)) is 1/y * y', because of the chain rule.


David Witten

(Not) The Second Derivative Test

Proving the limit of a quadratic