questão de derivadas?

targieli-d

y=raiz de (3x-x^2) determine a derivada de 2ª ordem

me ajudaaa

obg

maria-descartes

Veja:

y=(3x-x²)^1/2

y'=(3-2x)/2√(3x-x²)

y"=[(-2)2√(3x-x²)-(3-2x)2.1/2(3x-x²)^-1/2]/[2√(3x-x²)]²

y"=[-4√(3x-x²)-(3-2x)/√(3x-x²)]/[4(3x-x²)]

y"=[-4(3x-x²)-3+2x]/√(3x-x²)][4(3x-x²)]

y"=[-12x+4x²-3+2x]/√(3x-x²)][4(3x-x²)]

y"=[4x²-10x-3]/√(3x-x²)][4(3x-x²)]

luiz-s

y = raiz de (3x-x^2) = (3x-x²)^1/2

y' = (1/2)*(3 - 2x) / [ (3x - x²)^1/2 ]

y' = (3 - 2x) / [ 2(3x - x²)^1/2 ]

e

y'' = (1/2)*{- 2[(3x - x²)^1/2 ] + (3 - 2x)*[(3 - 2x) / [2(3x - x²)^1/2 ]]} / (3x - x²)

y'' = (1/2)*{[ -2(3x - x²)^1/2 ] + [(3 - 2x)² / [2(3x - x²)^1/2 ]]} / (3x - x²)

y'' = (1/2)*{[ -2(3x - x²)^1/2 ]*[2(3x - x²)^1/2 ] + (3 - 2x)² } / [2(3x - x²)^3/2]

y'' = (1/2)*{[ -4(3x - x²)] + (3 - 2x)² } / [2(3x - x²)^3/2]

y'' = {[ -4(3x - x²)] + (3 - 2x)² } / [4(3x - x²)^3/2]

:.

math-girl

y=(3x-x²)^1/2

Fazendo u = 3x-x² , temos:

y = u^1/2

(Aplicando a regra da cadeia)

y' = 1/2*u^(-1/2)*du/dx

y' = 1/2*(3x-x² )^(-1/2)*(3-2x)

y' = (3x-x² )^(-1/2)*(3-2x)/2

y' = (3x-x² )^(-1/2)*(3/2-x)

Fazendo v = (3x-x² )^(-1/2) e w = (3/2-x), temos:

(Aplicando a regra do produto)

y" = v w' + v' w

y" = (3x-x² )^(-1/2) (-1) -1/2 * (3x-x² )^(-3/2)*(3-2x)* (3/2-x)

y" = -[(3x-x² )^(-1/2)] -1/2 *[ (3x-x² )^(-3/2) (9/2-3x-6x+2x²)]

y" = -[(3x-x² )^(-1/2)] +[ (3x-x² )^(-3/2) (-9/4+9/2x-x²)]

Kisses

=**

sandrolsc2

assim:

raiz de (3x-x^2) é igual a = (3x-x^2)^1/2

1/2 *(3x-x^2) ^-1/2 * a derivada interna

1/2 *(3x-x^2)^ -1/2* (3 -2x)

(3-2x)/2 *(3x-x^2)^ -1/2

eu deixaria assim