Answer:
[tex]f(x,y,z)=ye^{xz}+C[/tex]
Step-by-step explanation:
We can write the given expression as :
[tex]\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k[/tex]
As given, f = ∇f.
∇f = [tex]\dfrac{\partial f}{\partial x}i[/tex] + [tex]\dfrac{\partial f}{\partial y}j[/tex] +[tex]\dfrac{\partial f}{\partial z}k [/tex]
We can write the partial derivative with respect to x, y and z.
[tex]\dfrac{\partial f}{\partial x}=yze^{xz}[/tex] ___(Equation 1)
[tex]\dfrac{\partial f}{\partial y}=e^{xz}[/tex] ______(Equation 2)
[tex]\dfrac{\partial f}{\partial z}=xye^{xz}[/tex] ______(Equation 3)
Take equation 2 and integrate with respect to y,
[tex]\dfrac{\partial f}{\partial y}=e^{xz}[/tex]
[tex]f(x,y,z)=ye^{xz}+a(x,z)[/tex] ----------Equation 4
Derivate both sides w.r.t x , we get :
[tex]\frac{d}{dx}(yze^{xz})=yze^{xz}+\dfrac{\partial a}{\partial x}[/tex]
or
[tex]\dfrac{\partial a}{\partial x}=0[/tex]
integrate
a(x,z)=b(z)
put in equation 4 ,
we get :
[tex]f(x,y,z)=ye^{xz}+b(z)[/tex]
take derivative wrt z
[tex]\frac{d}{dz} (ye^{xz}+b(z))\impliesxye^{xz}=xye^{xz}+\frac{db}{dz}[/tex]
we can take here:
[tex]\frac{db}{dz} = 0[/tex]
integrate:
[tex]\int\ {\frac{db}{dz} } \, =\int0[/tex]
b(z) = C
The function can be written as :
from equation 4 :
[tex]f(x,y,z)=ye^{xz}+C[/tex]
Where C is a constant.
