Question

enncontrar la primera derivada x^2 +sen^-1 y =ye^x

Answer 1

$\dfrac{d}{dx}(x^2+\sin^{-1}y)=\dfrac{d}{dx}(ye^x)\\ \\ \\ \dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(\sin^{-1}y)=\dfrac{dy}{dx}\cdot e^x+y\cdot\dfrac{d}{dx}(e^x)\\ \\ \\ 2x+\dfrac{d}{dy}(\sin^{-1}y)\cdot \dfrac{dy}{dx}=y'e^x+ye^x\\ \\ \\ 2x+\dfrac{y'}{\sqrt{1-y^2}}=y'e^x+ye^x\\ \\ \\ y'\left(\dfrac{1}{\sqrt{1-y^2}}-e^x\right)=ye^x-2x\\ \\ \\ \\ y'=\dfrac{ye^x-2x}{\frac{1}{\sqrt{1-y^2}}-e^x}$