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Exercise 8.4 · Q2

Q.For each of the following functions find the fx,fyf_x,f_y, and show that fxy=fyxf_{xy}=f_{yx}.

(i) f(x,y)=3xy+sin⁡xf(x,y)=\dfrac{3x}{y+\sin x}
(ii) f(x,y)=tan⁡−1(xy)f(x,y)=\tan^{-1}\left(\dfrac{x}{y}\right)
(iii) f(x,y)=cos⁡(x2−3xy)f(x,y)=\cos(x^2-3xy)
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✓ Free question

For each ff, find fx,fyf_x,f_y by the quotient/chain rule, then differentiate fxf_x w.r.t. yy and fyf_y w.r.t. xx separately and verify the two mixed partials agree (Clairaut's Theorem).

Part (i): f(x,y)=3xy+sin⁡xf(x,y)=\dfrac{3x}{y+\sin x}.

fx=3(y+sin⁡x)−3xcos⁡x(y+sin⁡x)2=3y+3sin⁡x−3xcos⁡x(y+sin⁡x)2f_x=\dfrac{3(y+\sin x)-3x\cos x}{(y+\sin x)^2}=\dfrac{3y+3\sin x-3x\cos x}{(y+\sin x)^2}. fy=−3x(y+sin⁡x)2\quad f_y=\dfrac{-3x}{(y+\sin x)^2}.

fxy=∂∂y[3y+3sin⁡x−3xcos⁡x(y+sin⁡x)2]f_{xy}=\dfrac{\partial}{\partial y}\left[\dfrac{3y+3\sin x-3x\cos x}{(y+\sin x)^2}\right]: with numerator N=3y+3sin⁡x−3xcos⁡xN=3y+3\sin x-3x\cos x (Ny=3N_y=3) and denominator D=(y+sin⁡x)2D=(y+\sin x)^2 (Dy=2(y+sin⁡x)D_y=2(y+\sin x)),

fxy=3D−N⋅2(y+sin⁡x)D2=3(y+sin⁡x)−2N(y+sin⁡x)3=3(y+sin⁡x)−2(3y+3sin⁡x−3xcos⁡x)(y+sin⁡x)3=6xcos⁡x−3y−3sin⁡x(y+sin⁡x)3.f_{xy}=\frac{3D-N\cdot2(y+\sin x)}{D^2}=\frac{3(y+\sin x)-2N}{(y+\sin x)^3}=\frac{3(y+\sin x)-2(3y+3\sin x-3x\cos x)}{(y+\sin x)^3}=\frac{6x\cos x-3y-3\sin x}{(y+\sin x)^3}.

fyx=∂∂x[−3x(y+sin⁡x)2]=−3(y+sin⁡x)2−(−3x)⋅2(y+sin⁡x)cos⁡x(y+sin⁡x)4=−3(y+sin⁡x)+6xcos⁡x(y+sin⁡x)3=6xcos⁡x−3y−3sin⁡x(y+sin⁡x)3.f_{yx}=\dfrac{\partial}{\partial x}\left[\dfrac{-3x}{(y+\sin x)^2}\right]=\dfrac{-3(y+\sin x)^2-(-3x)\cdot2(y+\sin x)\cos x}{(y+\sin x)^4}=\dfrac{-3(y+\sin x)+6x\cos x}{(y+\sin x)^3}=\dfrac{6x\cos x-3y-3\sin x}{(y+\sin x)^3}.

Both equal 3(2xcos⁡x−y−sin⁡x)(y+sin⁡x)3\dfrac{3(2x\cos x-y-\sin x)}{(y+\sin x)^3}: fxy=fyxf_{xy}=f_{yx}, confirmed.

Part (ii): f(x,y)=tan⁡−1 ⁣(xy)f(x,y)=\tan^{-1}\!\left(\dfrac xy\right).

fx=11+(x/y)2⋅1y=yx2+y2f_x=\dfrac{1}{1+(x/y)^2}\cdot\dfrac1y=\dfrac{y}{x^2+y^2}. fy=11+(x/y)2⋅(−xy2)=−xx2+y2\quad f_y=\dfrac{1}{1+(x/y)^2}\cdot\left(-\dfrac{x}{y^2}\right)=\dfrac{-x}{x^2+y^2}.

fxy=∂∂y[yx2+y2]=(x2+y2)−y(2y)(x2+y2)2=x2−y2(x2+y2)2f_{xy}=\dfrac{\partial}{\partial y}\left[\dfrac{y}{x^2+y^2}\right]=\dfrac{(x^2+y^2)-y(2y)}{(x^2+y^2)^2}=\dfrac{x^2-y^2}{(x^2+y^2)^2}.

fyx=∂∂x[−xx2+y2]=−(x2+y2)−(−x)(2x)(x2+y2)2=−(x2+y2)+2x2(x2+y2)2=x2−y2(x2+y2)2f_{yx}=\dfrac{\partial}{\partial x}\left[\dfrac{-x}{x^2+y^2}\right]=\dfrac{-(x^2+y^2)-(-x)(2x)}{(x^2+y^2)^2}=\dfrac{-(x^2+y^2)+2x^2}{(x^2+y^2)^2}=\dfrac{x^2-y^2}{(x^2+y^2)^2}.

Both equal x2−y2(x2+y2)2\dfrac{x^2-y^2}{(x^2+y^2)^2}: fxy=fyxf_{xy}=f_{yx}, confirmed.

Part (iii): f(x,y)=cos⁡(x2−3xy)f(x,y)=\cos(x^2-3xy).

fx=−sin⁡(x2−3xy)(2x−3y)f_x=-\sin(x^2-3xy)(2x-3y). fy=−sin⁡(x2−3xy)(−3x)=3xsin⁡(x2−3xy)\quad f_y=-\sin(x^2-3xy)(-3x)=3x\sin(x^2-3xy).

fxy=∂∂y[−(2x−3y)sin⁡(x2−3xy)]f_{xy}=\dfrac{\partial}{\partial y}\big[-(2x-3y)\sin(x^2-3xy)\big]: by the product rule with u=−(2x−3y), w=sin⁡(x2−3xy)u=-(2x-3y),\,w=\sin(x^2-3xy): uy=3u_y=3, wy=cos⁡(x2−3xy)(−3x)w_y=\cos(x^2-3xy)(-3x), so

fxy=3sin⁡(x2−3xy)+[−(2x−3y)]⋅[−3xcos⁡(x2−3xy)]=3sin⁡(x2−3xy)+3x(2x−3y)cos⁡(x2−3xy).f_{xy}=3\sin(x^2-3xy)+\big[-(2x-3y)\big]\cdot\big[-3x\cos(x^2-3xy)\big]=3\sin(x^2-3xy)+3x(2x-3y)\cos(x^2-3xy).

fyx=∂∂x[3xsin⁡(x2−3xy)]=3sin⁡(x2−3xy)+3xcos⁡(x2−3xy)(2x−3y)f_{yx}=\dfrac{\partial}{\partial x}\big[3x\sin(x^2-3xy)\big]=3\sin(x^2-3xy)+3x\cos(x^2-3xy)(2x-3y) (product rule).

Both equal 3sin⁡(x2−3xy)+3x(2x−3y)cos⁡(x2−3xy)3\sin(x^2-3xy)+3x(2x-3y)\cos(x^2-3xy): fxy=fyxf_{xy}=f_{yx}, confirmed.

✓Final answer

  1. fx=3y+3sin⁡x−3xcos⁡x(y+sin⁡x)2, fy=−3x(y+sin⁡x)2, fxy=fyx=3(2xcos⁡x−y−sin⁡x)(y+sin⁡x)3f_x=\dfrac{3y+3\sin x-3x\cos x}{(y+\sin x)^2},\ f_y=\dfrac{-3x}{(y+\sin x)^2},\ f_{xy}=f_{yx}=\boxed{\dfrac{3(2x\cos x-y-\sin x)}{(y+\sin x)^3}}
  2. fx=yx2+y2, fy=−xx2+y2, fxy=fyx=x2−y2(x2+y2)2f_x=\dfrac{y}{x^2+y^2},\ f_y=\dfrac{-x}{x^2+y^2},\ f_{xy}=f_{yx}=\boxed{\dfrac{x^2-y^2}{(x^2+y^2)^2}}
  3. fx=−(2x−3y)sin⁡(x2−3xy), fy=3xsin⁡(x2−3xy), fxy=fyx=3sin⁡(x2−3xy)+3x(2x−3y)cos⁡(x2−3xy)f_x=-(2x-3y)\sin(x^2-3xy),\ f_y=3x\sin(x^2-3xy),\ f_{xy}=f_{yx}=\boxed{3\sin(x^2-3xy)+3x(2x-3y)\cos(x^2-3xy)}

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