Solutions to Examples on Partial Derivatives ∂f ∂f = 3; = 4. ∂x ∂y ∂f ∂f f (x, y) = xy 3 + x2 y 2 ; = y 3 + 2xy 2 ; = 3xy 2 + 2x2 y. ∂x ∂y ∂f ∂f f (x, y) = x3 y + ex ; = 3x2 y + ex ; = x3 . ∂x ∂y ∂f ∂f f (x, y) = xe2x+3y ; = 2xe2x+3y + e2x+3y ; = 3xe2x+3y . ∂x ∂y x−y . f (x, y) = x+y ∂f x + y − (x − y) 2y = = ; ∂x (x + y)2 (x + y)2 ∂f −(x + y) − (x − y) 2x = =− . 2 ∂y (x + y) (x + y)2 1. (a) f (x, y) = 3x + 4y; (b) (c) (d) (e) (f) f (x, y) = 2x sin(x2 y). ∂f = 2x . cos(x2 y) . 2xy + 2 sin(x2 y) = 4x2 y cos(x2 y) + 2 sin(x2 y); ∂x ∂f = 2x . cos(x2 y) . x2 = 2x3 cos(x2 y). ∂y 2. f (x, y, z) = x cos z + x2 y 3 ez . ∂f = cos z + 2xy 3 ez , ∂x ∂f = 3x2 y 2 ez , ∂y ∂f = −x sin z + x2 y 3 ez . ∂z 1 3. (i) f (x, y) = x2 sin y + y 2 cos x. fx = 2x sin y − y 2 sin x; fy = x2 cos y + 2y cos x. fxx = 2 sin y − y 2 cos x; fyy = −x2 sin y + 2 cos x; fxy = 2x cos y − 2y sin x; fyx = 2x cos y − 2y sin x. So fxy = fyx . y ln x. x y 1 y y 1 fx = . − 2 ln x = 2 (1 − ln x); fy = ln x. x x x x x y 1 2y y fxx = 2 . − − 3 (1 − ln x) = 3 (2 ln x − 3); fyy = 0. x x x x 1 1 1 1 1 fxy = 2 (1 − ln x); fyx = . − 2 ln x = 2 (1 − ln x). x x x x x (ii) f (x, y) = So fxy = fyx . 4. 1 . + y2 ∂f ∂f ∂x ∂f ∂y = + . ∂t ∂x ∂t ∂y ∂t ∂f −2x −2r cos t ∂x = 2 = ; = −r sin t; 2 2 ∂x (x + y ) r4 ∂t ∂f −2y −2r sin t ∂y = 2 = ; = r cos t. 2 2 ∂y (x + y ) r4 ∂t ∂f 2r2 sin t cos t 2r2 sin t cos t Hence, = − = 0. ∂t r4 r4 ∂f ∂f ∂x ∂f ∂y = + . ∂r ∂x ∂r ∂y ∂r ∂f ∂f ∂x ∂y and are as above. = cos t, = sin t ∂x ∂y ∂r ∂r ∂f −2r cos2 t − 2r sin2 t −2(cos2 t + sin2 t) −2 Hence, = = = 3. ∂r r4 r3 r f (x, y) = x2 2 5. f (x, y) = x2 + xy − y 2 . (i) f (r, θ) = (r cos θ)2 + (r cos θ) (r sin θ) − (r sin θ)2 = r2 (cos2 θ + cos θ sin θ − sin2 θ). ∂f = 2r (cos2 θ + cos θ sin θ − sin2 θ). ∂r ∂f = r2 (−2 cos θ sin θ + cos θ cos θ − sin θ sin θ − 2 sin θ cos θ) ∂θ = r2 (cos2 θ − sin2 θ − 4 cos θ sin θ). (ii) ∂x = cos θ; ∂r ∂x ∂y ∂y = −r sin θ; = sin θ; = r cos θ. ∂θ ∂r ∂θ ∂f ∂f ∂x ∂f ∂y By the chain rule = + = (2x + y) cos θ + (x − 2y) sin θ ∂r ∂x ∂r ∂y ∂r = (2r cos θ + r sin θ) cos θ + (r cos θ − 2r sin θ) sin θ = 2r(cos2 θ + cos θ sin θ − sin2 θ). ∂f ∂f ∂x ∂f ∂y = + = (2x + y) (−r sin θ) + (x − 2y) r cos θ ∂θ ∂x ∂θ ∂y ∂θ = (2r cos θ + r sin θ) (−r sin θ) + (r cos θ − 2r sin θ) r cos θ = r2 (cos2 θ − sin2 θ − 4 cos θ sin θ). 6. f (x, y) = x3 y − y 3 x; ∂f ∂f ∂x ∂f ∂y = + ; ∂u ∂x ∂u ∂y ∂u u . v ∂f ∂f ∂x ∂f ∂y = + . ∂v ∂x ∂v ∂y ∂v x = uv; y= ∂f u u3 u3 = 3x2 y − y 3 = 3u2 v 2 − 3 = 3u3 v − 3 . ∂x v v v 3 2 3 u u 3u ∂f = x3 − 3y 2 x = 3 − 3 2 uv = u3 v 3 − . ∂y v v v ∂x ∂x ∂y 1 ∂y −u = v; = u; = ; = 2. ∂u ∂v ∂u v ∂v v ! ! 3u3 u3 3u v − 3 v + u3 v 3 − v v 3 3 u 3u = 3u3 v 2 − 2 + u3 v 2 − 2 = 4u3 v 2 − v v ∂f = ∂u 3 3 1 v 4u3 . v2 ! ! u3 3u3 −u 3u v − 3 u + u3 v 3 − v v v2 u4 3u4 2u4 = 3u4 v − 3 − u4 v + 3 = 2u4 v + 3 . v v v ∂f = ∂v 3 7. f (x, y, z) = 2y − sin(xz), x = 3t, y = et−1 , z = ln t. ∂f ∂t ∂f ∂x dx dt ∂f ∂t ∂f dx ∂f dy ∂f dz + + . ∂x dt ∂y dt ∂z dt ∂f ∂f = −z cos(xz); = 2; = −x cos(xz). ∂y ∂z dy dz 1 = 3; = et−1 ; = . dt dt t x cos(xz) = −3z cos(xz) + 2et−1 − t 3t cos(3t ln t) = −3 ln t cos(3t ln t) + 2et−1 − t = = −3 cos(3t ln t)(1 + ln t) + 2et−1 . 8. f (x, y) = x2 + xy + y 2 , x = uv, y = u/v. To show that ufu + vfv = 2xfx and ufu − vfv = 2yfy we need to find fu , fv , fx and fy . ∂f ∂f ∂x ∂f ∂y = + ; ∂u ∂x ∂u ∂y ∂u ∂f ∂f ∂x ∂f ∂y = + . ∂v ∂x ∂v ∂y ∂v 1 2u fu = (2x + y)(v) + (x + 2y)( ) = 2uv 2 + 2u + 2 . v v 2 2u −u fv = (2x + y)(u) + (x + 2y)( 2 ) = 2u2 v − 3 . v v u fx = 2x + y = 2uv + . So, 2xfx = 2uvfx = 4u2 v 2 + 2u2 . v 2u 4u2 2u fy = x + 2y = uv + . So, 2yfy = fx = 2u2 + 2 . v v v 2 2 2u 2u Now ufu + vfv = 2u2 v 2 + 2u2 + 2 + 2u2 v 2 − 2 v v = 4u2 v 2 + 2u2 = 2xfx as required, fu = fv = 4 and ufu − vfv = 2u2 v 2 + 2u2 + = 2u2 + 2u2 2u2 2 2 − 2u v + v2 v2 4u2 = 2yfy as required. v2 9. u(x, y) = ln(1 + xy 2 ). ∂u 1 y2 ∂2u −y 2 . y 2 y4 2 = . y = ; = = − . ∂x 1 + xy 2 1 + xy 2 ∂x2 (1 + xy 2 )2 (1 + xy 2 )2 ∂2u (1 + xy 2 ) . 2y − 2xy . y 2 2y = = . ∂y∂x (1 + xy 2 )2 (1 + xy 2 )2 ∂2u ∂2u 2y 4 2y 4 Hence 2 2 + y 3 =− + = 0. ∂x ∂y∂x (1 + xy 2 )2 (1 + xy 2 )2 10. u(x, y) = x2 cosh(xy 2 + 1). NOTE. d d (sinh x) = cosh x; (cosh x) = sinh x. dx dx ∂u = x2 sinh(xy 2 + 1) . y 2 + 2x cosh(xy 2 + 1) ∂x = x2 y 2 sinh(xy 2 + 1) + 2x cosh(xy 2 + 1) ∂u = x2 sinh(xy 2 + 1) . 2xy = 2x3 y sinh(xy 2 + 1) ∂y Hence 2x ∂u ∂u −y = 2x3 y 2 sinh(xy 2 +1)+4x2 cosh(xy 2 +1)−2x3 y 2 sinh(xy 2 +1) ∂x ∂y = 4x2 cosh(xy 2 + 1) = 4u. 11. Hence 1 ∂w = 2c ∂t 2x + 2ct ∂2w −4c2 = ∂t2 (2x + 2ct)2 ∂w 1 = 2 ∂x 2x + 2ct ∂2w −4 = 2 ∂x (2x + 2ct)2 2 ∂2w 2∂ w = c . ∂t2 ∂x2 5