| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Here's a good old Calculus question:
f(x) = ln square root(7-2X**2) <- 2x squared
find f'(x).
----------------------
now if f(x) = ln x then f'(x) = 1/x
How do you get rid of the square root??
Peter
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 577.1 | answer | GALLO::EKLUND | Dave Eklund | Fri Sep 12 1986 17:56 | 7 |
Use the identity:
ln (sqrt (g(x)) ) == 1/2 * ln (g(x))
This eliminates the square root. Chaining then makes the problem
trivial.
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| 577.2 | CLT::GILBERT | eager like a child | Sat Sep 13 1986 00:50 | 18 | |
Or, to be unsuave about this, we can apply the chain rule directly:
f(x) = ln (sqrt(7-2x^2))
d
f'(x) = -- (sqrt(7-2x^2)) / sqrt(7-2x^2)
dx
1 -1/2 d
= - (7-2x^2) -- (7-2x^2) / sqrt(7-2x^2)
2 dx
1 -1/2
= - (7-2x^2) (-4x) / sqrt(7-2x^2)
2
= -2x / (7-2x^2)
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