Question

URGENTE,ME AJUDEM
Responda usando a regra da cadeia z=2x+3y; x+t², y=5t?
z=x²y; x=3t+1,y=t²-1
z=3x/y; x=t,y=t²
z=x¹/2 y¹/3; x=2t,y=2t²
z=xy; x=e²t;y=e-³t
z=x+y/x-y; x= t³+1,y=1-t²

Answer 1

z = 2x+3y
x=t²
y=5t

z = 2(t²)+3(5t) = 2t² + 15t
z' = 2 . 2t + 15 = 4t + 15
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z=x²y; x=3t+1,y=t²-1 

z = (3t+1)².(t²-1) (regra do produto u'v+uv')
z' = 2.(3t+1).(3).(t²-1) + (3t+1)².(2t)
z' = 6(3t+1) + (9t²+6t+1)(2t)
z' = 12t+6+18t³+12t²+2t
z' = 18t³ + 12t²+14t+6
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z=3x/y; x=t,y=t² 

z = 3t/t² (regra do quociente: (u'v - uv')/v²)
z' = (3 . t² - 3t . 2t)/(t²)²
z' =  (3t² - 6t²)/t^4
z' = -3t²/t^4
z' = -3/t²
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z=x¹/2 y¹/3; x=2t,y=2t² 

$z = (2t)^{ \frac{1}{2} }. (2 t^{2} )^{ \frac{1}{3} }$ 
$z = \frac{1}{2} (2t)^{-\frac{1}{2}}.2.(2 t^{2} )^{ \frac{1}{3} } + (2t)^{ \frac{1}{2} }. \frac{1}{3} (2 t^{2} )^{ -\frac{2}{3} }.4t$
$z = \frac{(2 t^{2} )^{ \frac{1}{3} }}{(2t)^{ \frac{1}{2} }} + \frac{4t.(2t)^{ \frac{1}{2} }}{3.(2 t^{2} )^{ \frac{2}{3} } }$
$z = \frac{(2 t^{2} )^{ \frac{1}{3} }}{(2t)^{ \frac{1}{2} }} + \frac{4t.(2t)^{ \frac{1}{2} }}{3.(2 t^{2} )^{ \frac{2}{3} } } \\ \\ z = \frac{2^{ \frac{1}{3} } . t^{ \frac{2}{3} }}{2^{ \frac{1}{2} }.t^{ \frac{1}{2} }} + \frac{4t.2^{ \frac{1}{2} }.t^{ \frac{1}{2} }}{3.2 ^{ \frac{2}{3} }.t^{ \frac{4}{3} } } \\ \\ z = 2^{- \frac{1}{6} }.t^{ \frac{1}{6} } + \frac{4}{3}.2 ^{- \frac{1}{6} }. t ^{ \frac{1}{6} } \\ \\ z = \frac{10}{3}.(2 ^{- \frac{1}{6} }. t ^{ \frac{1}{6} })$
$z = \frac{10}{3}.(2 ^{- \frac{1}{6} }. t ^{ \frac{1}{6} }) \\ \\ z = \frac{10}{3}.( \frac{t ^{ \frac{1}{6} }}{2 ^{ \frac{1}{6} }} ) \\ \\ z = \frac{10}{3}. (\frac{t}{2}) ^{ \frac{1}{6} }$
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z=xy; x=e²t;y=e-³t 

$z = t e^{2}.t e^{-3} \\ \\ z' = e^{2} .t e^{-3} + t e^{2} . e^{-3} \\ \\ z' = t e^{-6} + t e^{-6} \\ \\ z' = 2t e^{-6}$
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z=x+y/x-y; x= t³+1,y=1-t²

$z = \frac{( t^{3} +1)+(1- t^{2} )}{( t^{3} +1)-(1- t^{2} )} = \frac{t^{3} +1+1- t^{2} }{ t^{3} +1-1- t^{2} } = \frac{t^{3} +2- t^{2} }{ t^{3} - t^{2} } \\ \\ z' = \frac{(3 t^{2}-2t ).( t^{3} - t^{2})-(t^{3} +2- t^{2}).(3 t^{2}-2t )}{ ( t^{3} - t^{2})^{2} } \\ \\ z' = \frac{(3 t^{2}-2t ).( t^{3} - t^{2})-(t^{3} +2- t^{2}).(3 t^{2}-2t )}{ ( t^{3} - t^{2})^{2} }$