How do you find the derivative of #4cosxsiny=1#?

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Answer 1 · 2017-09-15T23:38:43

#dy/dx = 1/4secxtanxcosy#

We have, using identities:

#siny = 1/(4cosx)#

#siny = 1/4secx#

Using implicit differentiation/known derivative, we get:

#cosy(dy/dx) = 1/4secxtanx#

Isolating the differential:

#dy/dx = 1/4secxtanxcosy#

Hopefully this helps!