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Partial Derivative (Calculus) question?
Here's the question/problem...
(NOTE, the "e" is that slanted/lowercase one that's equal to like 2.71...etc.)
f (x, t) = te^-(2xt)
...Find fx (x, t) and ft (x, t)
Show work. Fastest answer with correct answer gets best answer. Thank you!
2 Answers
- az_lenderLv 79 years agoFavorite Answer
The partial of f with respect to x is (chain rule)
(-2t) [te^(-2xt)] = -2t^2 e^(-2xt)
The partial of f with respect to t is (product rule)
(-2x) [te^(-2xt)] + (1)e^(-2xt)
= (1 - 2xt) e^(-2xt)
- Anonymous5 years ago
f(x,y) = x/(4 + y) = x(4 + y)^-a million so as that df/dx = a million/(4 + y) the place df/dx capacity differentiate f, in part, wrt x df/dy = -x/(4 + y)^2 the place df/dy capacity differentiate f, in part, wrt y
