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How do you differentiate #s=(1+sint)/(1+cost)#?

1 Answer

Feb 27, 2017

#(ds)/(dt) = ( 1+ cost + sint)/(1+cost)^2#

Explanation:

Using the quotient rule:

#(ds)/(dt) = ( (1+cost) d/(dt) (1+sint) - (1+sint) d/(dt) (1+cost))/(1+cost)^2#

#(ds)/(dt) = ( (1+cost) cost + (1+sint) sint)/(1+cost)^2#

#(ds)/(dt) = ( cost+cos^2t + sint+sin^2t)/(1+cost)^2#

#(ds)/(dt) = ( 1+ cost + sint)/(1+cost)^2#

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