CALCULAR DERIVADA IMPLÍCITA

(y+ xy')cos (xy)= cos(y)-y'xsen(y)

ycos(xy)+xy'cos(xy)=cos(y)-y'xsen(y)

xy'sen(y)+xy'cos(xy)=cos(y)-ycos(xy)

y'(xsen(y)+xcos(xy))=cosy-ycos(xy)

y'=cos(y)-ycos(xy)/xsen(y)+xcos(xy)

Seja:

\(\sin \left(xy\right)=x\cos \left(y\right)\)

Vamos derivar ambos os lados com relação a \(x\):

\(\frac{d}{dx}\left(\sin \left(xy\right)\right)=\frac{d}{dx}\left(x\cos \left(y\right)\right)\)

\(\frac{d}{dx}\left(\sin \left(xy\right)\right)=\frac{d}{dx}\left(x\cos \left(y\right)\right)\)

\(\frac{d}{dx}\left(x\cos \left(y\right)\right)=\cos \left(y\right)-x\sin \left(y\right)\frac{d}{dx}\left(y\right)\)

Assim:

\(\cos \left(xy\right)\left(y+x\frac{d}{dx}\left(y\right)\right)=\cos \left(y\right)-x\sin \left(y\right)\frac{d}{dx}\left(y\right)\)

Por conveniência, vamos escrever \(\frac{d}{dx}\left(y\right)\) como \(y'\):

\(\cos \left(xy\right)\left(y+xy^{'\:}\right)=\cos \left(y\right)-x\sin \left(y\right)y^{'\:}\)

Isolando o \(y'\):

\(y\cos \left(xy\right)+xy^{'\:}\cos \left(xy\right)=\cos \left(y\right)-x\sin \left(y\right)y^{'\:}\\ xy^{'\:}\cos \left(xy\right)=\cos \left(y\right)-x\sin \left(y\right)y^{'\:}-y\cos \left(xy\right)\\ xy^{'\:}\cos \left(xy\right)+x\sin \left(y\right)y^{'\:}=\cos \left(y\right)-y\cos \left(xy\right)\\ xy^{'\:}\left(\sin \left(y\right)+\cos \left(xy\right)\right)=\cos \left(y\right)-y\cos \left(xy\right)\\ y^{'\:}=\frac{\cos \left(y\right)-y\cos \left(xy\right)}{x\left(\cos \left(xy\right)+\sin \left(y\right)\right)}\)

Assim:

\(\frac{d}{dx}\left(\sin \left(xy\right)\right)=\frac{d}{dx}\left(x\cos \left(y\right)\right) \rightarrow \boxed{y^{'\:}=\frac{\cos \left(y\right)-y\cos \left(xy\right)}{x\left(\cos \left(xy\right)+\sin \left(y\right)\right)}}\)