Unizor - Derivatives - Reciprocal of a Function

Notes to a video lecture on http://www.unizor.com

Derivative Properties -
Reciprocal to a Function

Our purpose is to express the derivative of a reciprocal of a function in terms of its own derivative.

Assume that the real functionf(x) is defined anddifferentiable (that is, its derivative exist) on certain interval with a derivative f ^I(x).
Let's determine the derivative of its reciprocal
h(x) = 1/f(x) wherever f(x) ≠ 0.

The increment of function h(x)is
Δh(x) = h(x+Δx)−h(x) =
= 1/f(x+Δx) − 1/f(x) =
= [f(x)−f(x+Δx)]/[f(x+Δx)·f(x)]
= −Δf(x)/[f(x+Δx)·f(x)]

Next operations to find a derivative are: dividing the function increment Δh(x) by an increment of an argument Δxand going to a limit as Δx→0.

Δh(x)/Δx =
= [−Δf(x)/Δx)]/[f(x+Δx)·f(x)]

Since function f(x) is differentiable, there is a limit of Δf(x)/Δx as Δx→0. This limit isf ^I(x).

At the same time
f(x+Δx) → f(x)
as Δx→0.

Recall the properties of limits:
if two sequences have limits Land M, then their product has a limit L·M;
if two sequences have limits Land M ≠ 0, then their ratio has a limit L/M.

Therefore,
Δh(x)/Δx→−f ^I(x)/f²(x)

Hence,
h^I(x) = −f ^I(x)/f²(x)

It's called the Product Rule of differentiation.
Different forms of notation of this rule are:
(1) [1/f(x)] ^I = −f ^I(x)/f²(x)
(2) d[1/f(x)]/dx = −[df(x)/dx]·[1/f²(x)]

Examples

[sec(x)] ^I = [1/cos(x)] ^I =
= −[cos(x)] ^I/cos²(x) =
= sin(x)/cos²(x)

(d/dx)[tan(x)] =
= (d/dx)[sin(x)/cos(x)] =
= (d/dx){[sin(x)]·[1/cos(x)]} =
= (d/dx)[sin(x)]·[1/cos(x)] + sin(x)·(d/dx)[1/cos(x)] =
= cos(x)·[1/cos(x)] + sin(x)·sin(x)/cos²(x) =
= 1+tan²(x)

de^−x/dx =
= (d/dx)(1/e^x) =
= −[(d/dx)(e^x)]/[(e^x)²] =
= −e^x/[(e^x)²] =
= −1/e^x =
= −e^−x

D_x[f(x)/g(x)] =
= [D_xf(x)]·[1/g(x)] + f(x)·{D_x[1/g(x)]} =
= [D_xf(x)]·[1/g(x)] − f(x)·[D_xg(x)]/g²(x) =
= g(x)·[D_xf(x)]·/g²(x) − f(x)·[D_xg(x)]/g²(x) =
= [g(x)·f ^I(x)−f(x)·g ^I(x)]/g²(x)