Recordemos la derivada de un producto:
[tex]\boxed{\sf{\dfrac{d(u\cdot v)}{dx}=u\cdot\dfrac{dv}{dx}+v\cdot\dfrac{du}{dx}}}[/tex]
Aplicamos esta propiedad, solo que derivaremos respecto a "t"
[tex]\begin{array}{c}\sf{x\cdot y = 20}\\\\\sf{\dfrac{d}{dt}(x\cdot y) = \dfrac{d}{dt}(20)}\\\\\boxed{\sf{x\cdot \dfrac{dy}{dt} +y\cdot \dfrac{dx}{dt}=0}}\\\end{array}[/tex]
Del problema tenemos que [tex]\sf{\dfrac{dy}{dt}=10}[/tex] y [tex]\sf{x = 2}[/tex], pero nos falta el valor de "y", entonces
[tex]\sf{x\cdot y=20}\\\\\sf{y=\dfrac{20}{x}}\\\\\sf{y=\dfrac{20}{2}}\\\\\boldsymbol{\sf{y=10}}[/tex]
Ya con esto reemplazamos en la ecuación que está en el cuadrito de arriba
[tex]\begin{array}{c} \sf{x\cdot \dfrac{dy}{dt} +y\cdot \dfrac{dx}{dt}=0}\\\\\sf{(2)\cdot (10) +(10)\cdot \dfrac{dx}{dt}=0}\\\\\sf{20+10\cdot \dfrac{dx}{dt}=0}\\\\\sf{10 \dfrac{dx}{dt}=-20}\\\\\sf{\dfrac{dx}{dt}=-\dfrac{20}{10}}\\\\\boxed{\boxed{\boldsymbol{\sf{\dfrac{dx}{dt}=-2}}}}\end{array}[/tex]
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