Question

Calcule a derivada implícita (em relação a x) de: x.e^(x^2 + y^2) = 5

Answer 1

Olá!

$x \cdot e^{x^2 + y^2} = 5 \\\\ dx \cdot e^{x^2 + y^2} + x \cdot e^{x^2 + y^2} \cdot (2x \, dx + 2y \, dy) = 0 \\\\ x \cdot e^{x^2 + y^2} \cdot (2x \, dx + 2y \, dy) = - e^{x^2 + y^2} \, dx \;\; \div (e^{x^2 + y^2} \\\\ x \cdot (2x \, dx + 2y \, dy) = - dx \\\\ 2x^2 \, dx + 2xy \, dy = - dx \\\\ 2xy \, dy = - dx - 2x^2 \, dx \\\\ \boxed{\frac{dy}{dx} = - \frac{1 + 2x^2}{2xy}}$

Answer 2

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$\sf x\cdot e^{x^2+y^2}=5\\\sf e^{x^2+y^2}+x\cdot e^{x^2+y^2}\cdot\left(2x+2y\dfrac{dy}{dx}\right)=0\\\sf e^{x^2+y^2}+2x^2\cdot e^{x^2+y^2}+2xye^{x^2+y^2}\dfrac{dy}{dx}=0\\\sf 2xye^{x^2+y^2}\dfrac{dy}{dx}=-e^{x^2+y^2}\left[2x^2+1\right]\\\sf\dfrac{dy}{dx}=-\dfrac{\diagup\!\!\!\!\!e^{x^2+y^2}\left[2x^2+1\right]}{2xy\diagup\!\!\!\!\!\!e^{x^2+y^2}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\dfrac{dy}{dx}=-\dfrac{2x^2+1}{2xy}}}}}$