Question

derivada implicita das equçoes3y=e^xy

Answer 1

Temos o seguinte:

$3y = e^{xy}$

Logo:

$[3y]' = [e^{xy}]' \\ \\ 3'*y + y'*3 = e^{xy}*[xy]' \\ \\ 3y' = e^{xy}*[x'*y + y'*x] \\ \\ 3y' = e^{xy}*[y +xy'] \\ \\ 3y' = e^{xy}*y + e^{xy}*xy' \\ \\ 3y' - e^{xy}*xy' = e^{xy}*y \\ \\ y'(3 - e^{xy}x) = e^{xy}y \\ \\ \boxed {y' = \dfrac{dy}{dx} = \dfrac{e^{xy}y}{3 - e^{xy}x} }$

Answer 2

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$\sf3y=e^{xy}\\\sf 3\dfrac{dy}{dx}=e^{xy}\cdot\left(y+x\dfrac{dy}{dx}\right)\\\sf 3\dfrac{dy}{dx}=e^{xy}\cdot y+e^{xy}\cdot x\dfrac{dy}{dx}\\\sf3\dfrac{dy}{dx}-e^{xy}\cdot x\dfrac{dy}{dx}=ye^{xy}\\\sf\dfrac{dy}{dx}(3-xe^{xy})=ye^{xy}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\dfrac{dy}{dx}=\dfrac{ye^{xy}}{3-xe^{xy}}}}}}$