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Find dy/dx of xy=cot^3(y+x)^2
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Implicit D!
You need: what is the derivative of \[\frac{d}{dx}\cot(x)\]
-csc^2(x)
Ok so what is \[\frac{d}{dx}xy\] using implicit D? You need the product rule here.
x+y?
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well the product rule gives you (ab)' = ab' + a'b
You can use the chain rule on (x+y)^2 too.
x+y
Oh some angle identity, eh?
ok, so implicit D would have \[\frac{d}{dx}xy = x\frac{d}{dx}y + y\frac{d}{dx}x\] which would then be just \[\frac{d}{dx}xy = x\frac{d}{dx}y + y\] or in short form \[\frac{d}{dx}xy = xy^{\prime} + y\]
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Implicit D is actually the chain rule but with variables. We don't know what y(x).
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