Skip to content
Exercise 8.4 · Q3

Q.If U(x,y,z)=x2+y2xy+3z2yU(x,y,z)=\dfrac{x^2+y^2}{xy}+3z^2y, find ∂U∂x,∂U∂y\dfrac{\partial U}{\partial x},\dfrac{\partial U}{\partial y}, and ∂U∂z\dfrac{\partial U}{\partial z}.

Puducherry TnboardTextbookSubjectiveImportance★★★★★
28% · 28/99 Questions
✓ Free question

Split the fraction x2+y2xy\dfrac{x^2+y^2}{xy} into xy+yx\dfrac xy+\dfrac yx first, which makes each partial derivative a routine power-rule computation.

Step 1. Simplify UU. U(x,y,z)=x2+y2xy+3z2y=x2xy+y2xy+3z2y=xy+yx+3z2yU(x,y,z)=\dfrac{x^2+y^2}{xy}+3z^2y = \dfrac{x^2}{xy}+\dfrac{y^2}{xy}+3z^2y = \dfrac xy+\dfrac yx+3z^2y.

Step 2. Differentiate w.r.t. xx (treating y,zy,z as constants): ∂∂x(xy)=1y\dfrac{\partial}{\partial x}\left(\dfrac xy\right)=\dfrac1y; ∂∂x(yx)=−yx2\dfrac{\partial}{\partial x}\left(\dfrac yx\right)=-\dfrac{y}{x^2}; ∂∂x(3z2y)=0\dfrac{\partial}{\partial x}(3z^2y)=0.

Ux=1y−yx2.U_x = \frac1y-\frac{y}{x^2}.

Step 3. Differentiate w.r.t. yy (treating x,zx,z as constants): ∂∂y(xy)=−xy2\dfrac{\partial}{\partial y}\left(\dfrac xy\right)=-\dfrac{x}{y^2}; ∂∂y(yx)=1x\dfrac{\partial}{\partial y}\left(\dfrac yx\right)=\dfrac1x; ∂∂y(3z2y)=3z2\dfrac{\partial}{\partial y}(3z^2y)=3z^2.

Uy=−xy2+1x+3z2.U_y = -\frac{x}{y^2}+\frac1x+3z^2.

Step 4. Differentiate w.r.t. zz (treating x,yx,y as constants): only 3z2y3z^2y depends on zz.

Uz=6zy.U_z = 6zy.

✓Final answer

Ux=1y−yx2U_x=\boxed{\dfrac1y-\dfrac{y}{x^2}},  Uy=−xy2+1x+3z2\ U_y=\boxed{-\dfrac{x}{y^2}+\dfrac1x+3z^2},  Uz=6zy\ U_z=\boxed{6zy}

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.