1)Find dy/dt if y = sin 2t + In 3t .
2)Differentiate In √(1+x)/(1-x) with respect to x.
Differentiation!!!
2010-05-11 10:03 pm
回答 (2)
2010-05-11 10:16 pm
2010-05-11 10:18 pm
1. y = sin2t + ln(3t)
Differentiate both sides with respect to t
dy/dt = d(sin2t)/d(2t) X d(2t)/dt + d(ln3t)/d(3t) X d(3t)/dt
= 2cos2t + 3/t
2010-05-11 14:18:14 補充：
2. ln√(1 + x)/(1 - x)
= 1/2 ln(1 + x)/(1 - x) = 1/2 [ln(1 + x) - ln(1 - x)]
This will be easier for our calculation
Differentiate with respect to x
d/dx {1/2 [ln(1 + x) - ln(1 - x)]}
= 1/2 [1/(1 + x) + 1/(1 - x)]
= 1/2 [(1 - x) + (1 + x)]/(1 + x)(1 - x)
= 1/(1 - x^2)
Differentiate both sides with respect to t
dy/dt = d(sin2t)/d(2t) X d(2t)/dt + d(ln3t)/d(3t) X d(3t)/dt
= 2cos2t + 3/t
2010-05-11 14:18:14 補充：
2. ln√(1 + x)/(1 - x)
= 1/2 ln(1 + x)/(1 - x) = 1/2 [ln(1 + x) - ln(1 - x)]
This will be easier for our calculation
Differentiate with respect to x
d/dx {1/2 [ln(1 + x) - ln(1 - x)]}
= 1/2 [1/(1 + x) + 1/(1 - x)]
= 1/2 [(1 - x) + (1 + x)]/(1 + x)(1 - x)
= 1/(1 - x^2)
收錄日期: 2021-04-13 17:14:16
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