BCA2nd sem Partial Differentiation of Z=log(x^2+y^2),Z=F(Y/X),Z=tan Inverse (y/x),U=log(x^2+y^2)/x+y

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Problems on Partial Differentiation- U=log(x^2+y^2+z^2) , U=tan^-1(2xy/x^2-y^2), Z=f(x+ay) +Q(x-ay). Problems on Partial Differentiation- U=log(x^2+y^2+z^2) , U=tan^-1(2xy/x^2-y^2), Z=f(x+ay) +Q(x-ay).
Partial Differentiation | Multivariable Calculus | z(x + y)=x²+y², (∂z/∂x−∂z/∂y)²=4(1−∂z/∂x −∂z/∂y) Partial Differentiation | Multivariable Calculus | z(x + y)=x²+y², (∂z/∂x−∂z/∂y)²=4(1−∂z/∂x −∂z/∂y)
z = log ( x^2 + xy + y^2 ). x ∂z/∂x+y ∂z/∂y = 2 .If u=e^(x^2+ y^2 ) , ∂u/∂x + y ∂u/∂y = 2u log u . z = log ( x^2 + xy + y^2 ). x ∂z/∂x+y ∂z/∂y = 2 .If u=e^(x^2+ y^2 ) , ∂u/∂x + y ∂u/∂y = 2u log u .
Partial Differentiation | u = log(tan x + tan y + tan z), sin2x(∂u/∂x)+sin2y(∂u/∂y)+sin2z(∂u/∂z)=2 Partial Differentiation | u = log(tan x + tan y + tan z), sin2x(∂u/∂x)+sin2y(∂u/∂y)+sin2z(∂u/∂z)=2
Partial Differentiation || 𝒛=𝒕𝒂𝒏^(−𝟏) (𝒚⁄𝒙) || VTU maths || Dr Prashant Patil Partial Differentiation || 𝒛=𝒕𝒂𝒏^(−𝟏) (𝒚⁄𝒙) || VTU maths || Dr Prashant Patil
Partial Differentiation | Multivariable Calculus | u = log√(x² + y² + z²) Partial Differentiation | Multivariable Calculus | u = log√(x² + y² + z²)
Partial Derivatives - Multivariable Calculus Partial Derivatives - Multivariable Calculus
Find the general integral of the PDE (x^2-y^2-yz)p+(x^2-y^2-zx)q=z(x-y) Find the general integral of the PDE (x^2-y^2-yz)p+(x^2-y^2-zx)q=z(x-y)