Differentiate #s=(1+sint)/(1+tant)# using the quotient rule?

Question

The answer is

#(cost(1+tant)-sec^2t(1+sint))/(1+tant)^2#

1100 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-05T05:12:58

Please see below.

According to Quotient rule, if #s(t)=(g(t))/(h(t))#

then #(ds)/(dt)=((dg)/(dt)xxh(t)-(dh)/(dt)xxg(t))/(h(t))^2#

Here in #s(t)=(1+sint)/(1+tant)#,

#g(t)=1+sint# and #(dg)/(dt)=cost# and

#h(t)=1+tant# and #(dh)/(dt)=sec^2t#

Hence #(ds)/(dt)=(cost(1+tant)-sec^2t(1+sint))/(1+tant)^2#