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gaellensah
@gaellensah
January 2021
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Bsr j'ai besoin d'aide en math s'il vous plaît je ne comprends pas mon exercice
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Bonjour,
1) a) dx(t)/dt = ( 3t²-t )'( 2-√t ) + ( 3t²-t )( 2-√t )'
= ( 6t-1 )( 2-√t ) + ( 3t²-t ) × ( -1/2√t )
= 12t - 6t√t - 2 +√t -3t√t/2 + √t/2
= 12t - 9t√t + 1,5√t - 2
b) dx(0)/dt n'est pas définie car 1/2√t n'est pas définie en 0
c) t→0, lim dx(t)/dt = -2
2) a) dg(x;y)/dx = 2( x-y ) + 2(x+y) = 4x ( on dérive selon x donc y est considéré comme constante )
dg(x;y)/dy = -2(x-y) + 2(x+y) = 4y
b) dh(a;b)/da = ( (5/b³) - a²/b² )' = -2a/b²
dh(a;b)/db =
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Verified answer
Bonjour,1) a) dx(t)/dt = ( 3t²-t )'( 2-√t ) + ( 3t²-t )( 2-√t )'
= ( 6t-1 )( 2-√t ) + ( 3t²-t ) × ( -1/2√t )
= 12t - 6t√t - 2 +√t -3t√t/2 + √t/2
= 12t - 9t√t + 1,5√t - 2
b) dx(0)/dt n'est pas définie car 1/2√t n'est pas définie en 0
c) t→0, lim dx(t)/dt = -2
2) a) dg(x;y)/dx = 2( x-y ) + 2(x+y) = 4x ( on dérive selon x donc y est considéré comme constante )
dg(x;y)/dy = -2(x-y) + 2(x+y) = 4y
b) dh(a;b)/da = ( (5/b³) - a²/b² )' = -2a/b²
dh(a;b)/db =