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**Foxlion** Problem:

$\displaystyle 4cosxsiny = 1$

Solution attempt:

create differentiation schema;

$\displaystyle siny[\frac{d}{dx}4(cosx)+4\frac{d}{dx}(cosx)] + 4cosx[\frac{d}{dx}siny]=\frac{d}{dx}1$

apply differentiation schema;

$\displaystyle siny[-4sinx]+4cosx[cosyy'] = 0$

simplify;

$\displaystyle -4sinxsiny+4cosxcosyy' = 0$

add $\displaystyle 4sinxsiny$ to both sides;

$\displaystyle 4cosxcosyy' = 4sinxsiny$

divide both sides by $\displaystyle 4cosxcosyy'$;

$\displaystyle y' = \frac{4sinxsiny}{4cosxcosy}$

apply trig identity $\displaystyle \frac{sinx}{cosx} = tanx$ and simplify;

$\displaystyle y' = tanxtany$

Answer has been verified, thank you all.