calculo função derivavél

por vinicius paiva duarte Sáb 06 Set 2014, 19:21

Seja f: R----->R uma função derivável, tal que f(5)=2; e f '(5)=3. Se z(x,y)=f((5)x^2+(-2)y), então o valor da expressão

z(1,0)+z_x(1,0)+z_y(1,0)

é igual a:

por Man Utd Dom 07 Set 2014, 12:01

Olá

Perceba:

$\\\\\\ z(x,y)=f(5x^2-2y)$

logo:

$\\\\\\ z(1,0)=f(5)=2$

$\\\\\\ z(x,y)=f(5x^2-2y) \\\\\\ z(x,y)_{x}=f^{\prime}(5x^2-2y)*(5x^2-2y)^{\prime} \\\\\\ z(x,y)_{x}=f^{\prime}(5x^2-2y)*10x^2 \\\\\\ z(1,0)_{x}=f^{\prime}(5)*10=30$

$\\\\\\ z(x,y)=f(5x^2-2y) \\\\\\ z(x,y)_{y}=f^{\prime}(5x^2-2y)*(5x^2-2y)^{\prime} \\\\\\ z(x,y)_{y}=f^{\prime}(5x^2-2y)*(-2) \\\\\\ z(1,0)_{y}=f^{\prime}(5)*(-2)=-6$

Agora some os valores.Confira com o gabarito por favor.