Let's use the quotient rule in a simple example. The quotient rule
tells us that:
d
dx
æ è
u
v
ö ø
=
du
dx
v-u
dv
dx
v2
In this example u will be 1, so we'll be finding the derivative of
[1/v], the reciprocal of v.
d
dx
æ è
1
v
ö ø
= ?
We're going to use the formula above. We know u = 1 and v=v, so we
still need to find [du/dx] and [dv/dx] before we can
apply the formula.
The derivative of a constant (like 1) is zero, so [du/dx] = 0. We don't know what v is, so we'll just write [dv/dx] = v¢. Plugging all this in to the quotient rule formula we get:
d
dx
æ è
1
v
ö ø
=
0 ·v-1v¢
v2
=
-v¢
v2
=
-v-2v¢
Now we have a general formula that lets us differentiate reciprocals!
Next, let's use this formula to see what happens when u = 1 and v = xn. Here again [du/dx] = 0 and now v¢ = [d/dx]xn = nxn-1.
d
dx
æ è
1
xn
ö ø
=
-v-2v¢
=
-(xn)-2(n xn-1)
=
-x-2n(n xn-1)
=
-nx-n-1
But [1/(xn)] = x-n, which is x to a power. We have a
rule for taking the derivative of x to a positive power; how does that compare to our new rule for the derivative of x to a negative power?
d
dx
x-n
=
-n x-n-1
This agrees with the formula [d/dx]xn = n xn-1, so the
quotient rule confirms that our rule for taking the derivative of
xn works even when n is negative.
File translated from
TEX
by
TTH,
version 3.82. On 19 Jul 2010, 18:05.