Iv also got a problem with derivative of trig functions, heres mine:
[sinInverse(x)/sqrt(1-x^2) dx] for range 0...0.5
am not sure how to solve that?
Here is one way.
Sorry but I forgot already those chain rule, quotient rule, whatever rule. I just keep on differentiating until all the variables are differentiated, etc. I must be following those said rules although those rules are not in my mind any more.
y = [(1 +cosX) / sinX]^(-1)
dy/dX = -1[(1 +cosX) / sinX]^(-2) *({sinX[-sinX] -(1 +cosX)[cosX]} / sin^2(X))
dy/dX = (-1 / [(1 +cosX) / sinX]^2) * ({-sin^2(X) -cosX -cos^2(X)} / sin^2(X))
dy/dX = -(sin^2(X) / (1 +cosX)^2) * -({sin^2(X) +cos^2(X) +cosX} / sin^2(X))
The sin^2(X) cancell out; the sin^2(X) +cos^2(X) is converted to 1; the multiplication of negatives turns to positive,
dy/dX = (1 +cosX) / (1 +cosX)^2
dy/dX = 1 / (1 +cosX) ------------------------answer.
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y = sec^2(pi*X)
Or,
y = [sec(pi*x)]^2
dy/dX = 2[sec(pi*X)] *[sec(pi*X)tan(pi*X)] *[pi]
dy/dX = (2pi)sec^2(pi*X)tan(pi*X) -----------------answer.