Given A ={(t^2)/2} -{[(1-t)^3]/5} find dA/dt thx
2007-07-30 01:56:05 · 4 answers · asked by bordemeriste b 1 in Science & Mathematics ➔ Mathematics
dA/dt = 2t/2 - 3[(-1)(1 - t)^2]/5 = t + (3/5)(1 - t)^2
When you have a function of the form f(x) = x^n, you get f'(x) = n*x^(n - 1). In the second term, there is an extra factor of (-1) due to the chain rule; we had (1 - t) raised to a power, and the derivative of (1 - t) is (-1).
2007-07-30 01:59:53 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
A = (1/2) t ² - (1/5) (1 - t) ³
dA / dt = t - (3/5) (1 - t) ² ( - 1)
dA /dt = t + (3/5) (1 - t) ²
2007-07-30 10:04:35 · answer #2 · answered by Como 7 · 1⤊ 0⤋
dA/Dt=t-3(1-t)^2(-1)/5
=t+3/5(1-t)^2
2007-07-30 09:02:42 · answer #3 · answered by Anonymous · 0⤊ 0⤋
= lamb.
2007-07-30 08:59:00 · answer #4 · answered by Bluebudgie21 5 · 0⤊ 3⤋