Question

ULTILIZE A REGRA DA CADEIA PARA DETERMINAR  AS DERIVADAS PARCIAIS INDICADAS
A)Z=X/Y X=RE ELEVADO A  ST, Y= RSE ELEVADO A T;


                             DZ/DR,DZ/DS,DZ/DT QUANDO R=1, S=2,T=0
ME AJUDEM PELO AMOR DE DEUS JA QUEBREI A CABEÇA E NAO CONSIGO FAZER PRECISO DISSO PRA HOJE ?

Answer 1

$Z = \frac{x}{y} \\\\\boxed{ \frac{dZ}{dx } =\frac{1}{y} } \\\\\\ \boxed{\frac{dZ}{dy} = \frac{-1}{y^2} }$

o enunciado diz 
$\boxed{X=R*e^{S*T}}\\\\\boxed{Y=R*S*e^{T}}$

e tambem diz que 
R =1 ; S =2 ; T =0

então 

$\frac{dZ}{dx } =\frac{1}{R*S*e^{T}}= \frac{1}{1*2*e^0}\to \boxed{ \frac{dZ}{dx}= \frac{1}{2} } \\\\\\ \frac{dZ}{dy} = \frac{-1}{y^2} }= \boxed{ \frac{dZ}{dy}= \frac{-1}{4} }$
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agora as derivadas
sabendo que
$\boxed{e^{(u)} = e^u * u'}$

derivando e x e substituindo 
R =1 ..S=2 ...T=0

$\frac{dx}{dR}=e^{S*t} =e^{2*0} = \boxed{ \frac{dx}{dR}=1}\\\\ \frac{dx}{dS}=R*e^{S*T}* (1*T)} = \boxed{ \frac{dx}{dS}=0}\\\\\\ \frac{dx}{dT}=R*e^{S*T} * (S*1) = \boxed{ \frac{dx}{dT}=2 }$

fazendo o mesmo processo para Y
$\frac{dy}{dR}=S*e^T = \boxed{ \frac{dy}{dR}= 2 } \\\\\\ \frac{dy}{dS}=R*e^{T} = \boxed{ \frac{dy}{ds}= 1} \\\\\\ \frac{dy}{dT}=R*S*e^{T} = \boxed{ \frac{dy}{dT}=2 }$
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agora calculando as derivadas parciais
$\frac{dZ}{dR}= \frac{dZ}{dR} * \frac{dx}{dR} + \frac{dZ}{dy} * \frac{dy}{dR} \\\\\frac{dZ}{dR}= \frac{1}{2}*1 + \frac{-1}{4}* 2 \\\\ \frac{dZ}{dR}= \frac{1}{2}- \frac{1}{2} \\\\\ \boxed{\frac{dZ}{dR} = 0 }$

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$\frac{dZ}{dS} = \frac{dZ}{dx} * \frac{dx}{dS} + \frac{dZ}{dy} * \frac{dy}{dS} \\\\ \frac{dZ}{dS} = \frac{1}{2}*0 + \frac{-1}{4} * 1 \\\\ \boxed{\frac{dZ}{dS} = \frac{-1}{4} }$

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$\frac{dZ}{dT} = \frac{dZ}{dx} * \frac{dx}{dT} + \frac{dZ}{dy} * \frac{dy}{dT} \\\\ \frac{dZ}{dT} = \frac{1}{2}*2 + \frac{-1}{4}*2 \\\\ \frac{dZ}{dT} =1 - \frac{1}{2} \\\\ \boxed{\frac{dZ}{dT} = \frac{1}{2} }$