Skip to content
Question 76 of 99

Q.If u=sin⁡−1(x+yx−y)⋅tan⁡(x3+y3x3−y3)u = \sin^{-1}\left(\dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} - \sqrt{y}}\right) \cdot \tan\left(\dfrac{x^3 + y^3}{x^3 - y^3}\right) then find x∂u∂x+y∂u∂yx\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y}.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2018Subjective· 6mImportance★★★★★
77% · 76/99 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Show each factor of u is homogeneous of degree 0 in x,y, so their product u is too, then apply Euler's theorem with n=0.

  1. Write u(x,y)=f(x,y)⋅g(x,y)u(x,y)=f(x,y)\cdot g(x,y) where f=sin⁡−1 ⁣(x+yx−y)f=\sin^{-1}\!\left(\dfrac{\sqrt x+\sqrt y}{\sqrt x-\sqrt y}\right) and g=tan⁡ ⁣(x3+y3x3−y3)g=\tan\!\left(\dfrac{x^3+y^3}{x^3-y^3}\right).
  2. Scale x→λx, y→λyx\to\lambda x,\ y\to\lambda y (λ>0\lambda>0) inside ff's argument: λx+λyλx−λy=λ(x+y)λ(x−y)=x+yx−y\dfrac{\sqrt{\lambda x}+\sqrt{\lambda y}}{\sqrt{\lambda x}-\sqrt{\lambda y}}=\dfrac{\sqrt\lambda(\sqrt x+\sqrt y)}{\sqrt\lambda(\sqrt x-\sqrt y)}=\dfrac{\sqrt x+\sqrt y}{\sqrt x-\sqrt y} — unchanged.
  3. So f(λx,λy)=f(x,y)f(\lambda x,\lambda y)=f(x,y), i.e. ff is homogeneous of degree 00.
  4. Scale inside gg's argument: (λx)3+(λy)3(λx)3−(λy)3=λ3(x3+y3)λ3(x3−y3)=x3+y3x3−y3\dfrac{(\lambda x)^3+(\lambda y)^3}{(\lambda x)^3-(\lambda y)^3}=\dfrac{\lambda^3(x^3+y^3)}{\lambda^3(x^3-y^3)}=\dfrac{x^3+y^3}{x^3-y^3} — also unchanged.
  5. So g(λx,λy)=g(x,y)g(\lambda x,\lambda y)=g(x,y), i.e. gg is homogeneous of degree 00. …

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.