Dada a função f(x, y, z) = sen(y + 2z) + ln(xyz), encontre ( ) + ( ) + ( )∂f/∂x ( ) + ( ) + ( )∂f/∂x ( ) + ( ) + ( )∂f/∂z + + + 2 cos(y + 2z)1/x ...

Para encontrar as derivadas parciais de f(x, y, z) = sen(y + 2z) + ln(xyz), vamos utilizar as regras de derivação: ∂f/∂x = ∂/∂x (sen(y + 2z) + ln(xyz)) ∂f/∂x = 0 + 1/x + 0 ∂f/∂x = 1/x ∂f/∂y = ∂/∂y (sen(y + 2z) + ln(xyz)) ∂f/∂y = cos(y + 2z) + 0 ∂f/∂y = cos(y + 2z) ∂f/∂z = ∂/∂z (sen(y + 2z) + ln(xyz)) ∂f/∂z = 2cos(y + 2z) + 0 ∂f/∂z = 2cos(y + 2z) Portanto, as respostas são: ( ) + ( ) + ( )∂f/∂x = 1/x + 0 + 0 ( ) + ( ) + ( )∂f/∂y = 0 + cos(y + 2z) + 0 ( ) + ( ) + ( )∂f/∂z = 0 + 0 + 2cos(y + 2z) + + + 2 cos(y + 2z)1/x 1/y 1/z = 2cos(y + 2z)/xyz + +   + 3 cos(y + 2z)1/x 1/y 1/z = 3cos(y + 2z)/xyz