Question

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

façam passo a passo pf . obrigado

Answer 1

$f(x,y,z)= sen(y+2z)+ln(xyz)$

lembrando que as derivadas:
$(C*u)'=C*u'\\\\\ [sen(u)]' = cos(u)*u'\\\\\ [ln(u)]'= \frac{1}{u}*u'$

aplicando isso:
∂f/∂x -> considera y e z como constantes

$\frac{\partial f}{\partial x} =0 + \frac{1}{xyz}*(xyz)' \\\\ \frac{\partial f}{\partial x} = \frac{1}{xyz}*(1yz) \\\\ \frac{\partial f}{\partial x} = \frac{yz}{xyz} = \frac{1}{x}$

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∂f/∂y -> considera x e z como constantes

$\frac{\partial f}{\partial y} = cos(y+2z)*(y+2z)'+ \frac{1}{xyz}* (xyz) '\\\\ \frac{\partial f}{\partial y} = cos(y+2z)*(1+0)+ \frac{1}{xyz}* (x*1*z)'\\\\ \frac{\partial f}{\partial y} = cos(y+2z)+ \frac{1}{y}$

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∂f/∂z -> considera x e y como constantes

$\frac{\partial f}{\partial y} = cos(y+2z)*(y+2z)'+ \frac{1}{xyz}* (xyz) '\\\\ \frac{\partial f}{\partial y} = cos(y+2z)*(0+2*1)+ \frac{1}{xyz}* (xy*1) \\\\ \frac{\partial f}{\partial y} = 2*cos(y+2z)+ \frac{1}{z}$

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$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} =\frac{1}{x}+cos(y+2z)+ \frac{1}{y}+2*cos(y+2z)+ \frac{1}{z} \\\\ \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} = 3cos(y+2z)+ \frac{1}{x}+ \frac{1}{y}+ \frac{1}{z} \\\\\boxed{\boxed{ \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} = 3cos(y+2z)+ \frac{yz+xz+xy}{xyz} }}$