Asked by Jovante
y=tan(x-y)
Answers
Answered by
Steve
y' = sec^2(x-y)*(1-y')
y'(1+sec^2(x-y)) = sec^2(x-y)
y' = sec^2(x-y)/(1+sec^2(x-y))
= 1 - 1/(sec^2(x-y)+1)
y'(1+sec^2(x-y)) = sec^2(x-y)
y' = sec^2(x-y)/(1+sec^2(x-y))
= 1 - 1/(sec^2(x-y)+1)
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