Question

derivada implicita de y^x=x^y

Answer 1

Resposta:

ver abaixo

Explicação passo-a-passo:

oi vamos lá, organizando a igualdade abaixo :

$y^x = x^y\Rightarrow \ln y^x =\ln x^y\Rightarrow x\cdot \ln y = y\cdot \ln x$   derivando agora :

$\ln y \ +\frac{x}{y}\cdot y'=y'\cdot \ln x \ +\frac{y}{x}$  agrupando os termos semelhantes :

$\frac{x}{y}\cdot y' - y'\cdot \ln x = \frac{y}{x} - \ln y \Rightarrow y'(\frac{x}{y} -\ln x) = \frac{y}{x}-\ln y\Rightarrow\\\\y' = \frac{\frac{y}{x}-\ln y}{\frac{x}{y} -\ln x}$

um abração

Answer 2

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$\sf y^x=x^y\\\sf e^{x\ell ny}=e^{y\ell nx}\\\sf e^{x\ell ny}\cdot\left(\ell ny+x\cdot\dfrac{1}{y}\dfrac{dy}{dx}\right)=e^{y\ell nx}\cdot\left(\dfrac{dy}{dx}\ell nx+y\cdot\dfrac{1}{x}\right)\\\sf e^{x\ell ny}\cdot\ell ny+\dfrac{x}{y}~\dfrac{dy}{dx}~e^{x\ell ny}=e^{y\ell nx}\cdot\dfrac{dy}{dx}\ell nx+e^{y\ell nx}\cdot\dfrac{y}{x}\\\sf \dfrac{x}{y}\cdot\dfrac{dy}{dx}e^{x\ell ny}-\ell nx\dfrac{dy}{dx}~e^{y\ell nx}=e^{y\ell nx}\cdot\dfrac{y}{x}-\ell ny\cdot e^{x\ell ny}$

$\sf\dfrac{dy}{dx}\left(\dfrac{x}{y}~e^{x\ell ny}-\ell nx \cdot e^{y\ell nx}\right)=e^{y\ell nx}\cdot\dfrac{y}{x}-\ell ny\cdot e^{x\ell ny}\\\sf\dfrac{dy}{dx}\left(\dfrac{x~e^{x\ell ny}-y~\ell nx\cdot e^{y\ell nx}}{y}\right)=\dfrac{y~e^{y\ell nx}-x~\ell ny\cdot e^{x\ell ny}}{x}\\\sf\dfrac{dy}{dx}=\dfrac{y\left[y~e^{y\ell nx}-x~\ell ny\cdot e^{x\ell ny}\right]}{x\left[x~e^{y\ell nx}-y~\ell n x\cdot e^{y\ell nx}\right]}$

$\huge\boxed{\boxed{\boxed{\boxed{\sf\dfrac{dy}{dx}=\dfrac{y\left[y\cdot x^y-x\ell ny\cdot y^x\right]}{x\left[x\cdot x^y-y\ell nx\cdot x^y\right]}}}}}$