For each f, find fx,fy by the quotient/chain rule, then differentiate fx w.r.t. y and fy w.r.t. x separately and verify the two mixed partials agree (Clairaut's Theorem).
Part (i): f(x,y)=y+sinx3x.
fx=(y+sinx)23(y+sinx)−3xcosx=(y+sinx)23y+3sinx−3xcosx. fy=(y+sinx)2−3x.
fxy=∂y∂[(y+sinx)23y+3sinx−3xcosx]: with numerator N=3y+3sinx−3xcosx (Ny=3) and denominator D=(y+sinx)2 (Dy=2(y+sinx)),
fxy=D23D−N⋅2(y+sinx)=(y+sinx)33(y+sinx)−2N=(y+sinx)33(y+sinx)−2(3y+3sinx−3xcosx)=(y+sinx)36xcosx−3y−3sinx.
fyx=∂x∂[(y+sinx)2−3x]=(y+sinx)4−3(y+sinx)2−(−3x)⋅2(y+sinx)cosx=(y+sinx)3−3(y+sinx)+6xcosx=(y+sinx)36xcosx−3y−3sinx.
Both equal (y+sinx)33(2xcosx−y−sinx): fxy=fyx, confirmed.
Part (ii): f(x,y)=tan−1(yx).
fx=1+(x/y)21⋅y1=x2+y2y. fy=1+(x/y)21⋅(−y2x)=x2+y2−x.
fxy=∂y∂[x2+y2y]=(x2+y2)2(x2+y2)−y(2y)=(x2+y2)2x2−y2.
fyx=∂x∂[x2+y2−x]=(x2+y2)2−(x2+y2)−(−x)(2x)=(x2+y2)2−(x2+y2)+2x2=(x2+y2)2x2−y2.
Both equal (x2+y2)2x2−y2: fxy=fyx, confirmed.
Part (iii): f(x,y)=cos(x2−3xy).
fx=−sin(x2−3xy)(2x−3y). fy=−sin(x2−3xy)(−3x)=3xsin(x2−3xy).
fxy=∂y∂[−(2x−3y)sin(x2−3xy)]: by the product rule with u=−(2x−3y),w=sin(x2−3xy): uy=3, wy=cos(x2−3xy)(−3x), so
fxy=3sin(x2−3xy)+[−(2x−3y)]⋅[−3xcos(x2−3xy)]=3sin(x2−3xy)+3x(2x−3y)cos(x2−3xy).
fyx=∂x∂[3xsin(x2−3xy)]=3sin(x2−3xy)+3xcos(x2−3xy)(2x−3y) (product rule).
Both equal 3sin(x2−3xy)+3x(2x−3y)cos(x2−3xy): fxy=fyx, confirmed.
✓Final answer
- fx=(y+sinx)23y+3sinx−3xcosx, fy=(y+sinx)2−3x, fxy=fyx=(y+sinx)33(2xcosx−y−sinx)
- fx=x2+y2y, fy=x2+y2−x, fxy=fyx=(x2+y2)2x2−y2
- fx=−(2x−3y)sin(x2−3xy), fy=3xsin(x2−3xy), fxy=fyx=3sin(x2−3xy)+3x(2x−3y)cos(x2−3xy)