Coached Exercise for Leibniz' Rule, step 1© 1999 by Karl Hahn |
|
The hint was that
e-x
f(x) =
x
is the product of two functions, g(x) and h(x). You were
to identify what they were.
1
g(x) = h(x) = e-x
x
If you had them vice versa, that's also correct.
Coached Exercise for Leibniz' Rule, step 2© 1999 by Karl Hahn |
|
Here you needed to take the first four derivatives of g(x) and h(x).
1
g'(x) = - h'(x) = -e-x
x2
1
g"(x) = 2 h"(x) = e-x
x3
1
g(3)(x) = -6 h(3)(x) = -e-x
x4
1
g(4)(x) = 24 h(4)(x) = e-x
x5
Coached Exercise for Leibniz' Rule, step 3© 1999 by Karl Hahn |
|
There is nothing left to do now but to apply Leibniz' Rule. You should have gotten from row 1 of the table:
f'(x) = g(x)h'(x) + g'(x)h(x)
1 1
= - e-x - e-x
x x2
From row 2 of the table:
f"(x) = g(x)h"(x) + 2g'(x)h'(x) + g"(x)h'(x)
1 2 2
= e-x + e-x + e-x
x x2 x3
From row 3 of the table:
f(3)(x) = g(x)h(3)(x) + 3g'(x)h"(x) + 3g"(x)h'(x) + g(3)(x)h(x)
1 3 6 6
= - e-x - e-x - e-x - e-x
x x2 x3 x4
From row 4 of the table:
f(4)(x) = g(x)h(4)(x) + 4g'(x)h(3)(x) + 6g"(x)h"(x) + 4g(3)(x)h'(x) + g(4)(x)h(x)
1 4 12 24 24
= e-x + e-x + e-x + e-x + e-x
x x2 x3 x4 x5
Notice that if you consider the the function itself as its zeroth derivative, then, for example, the
derivative numbers for each product in the fourth derivative add up to 4. The derivative
number for each product in the third derivative add up to 3. This will be the pattern
for any numbered derivative.
You can email me by clicking this button: