PROBLEMS ON CHAIN RULE IN PARTIAL DIFFERENTIATION

Suppose that W(x, y) is a function of two variables x, y having partial derivatives ∂W/∂x, ∂W/∂y. If both variables x, y are differentiable functions of a single variable t, then W is differentiable function t and 

Problem 1 :

If

w(x, y) = 6x3-3xy+2y2, x = es, y = cos s, s ∈ 

find dw/ds, and evaluate at s = 0.

Solution :

dw/ds = (∂w/∂x) (dx/ds) + (∂w/∂y) (dy/ds)

∂w/∂x = 6(3x2)-3(1)y+0

∂w/∂x = 18x2-3y

dx/ds = es

∂w/∂y = 0-3x(1)+2(2y)

∂w/∂y = -3x+4y

dy/ds = -sin s

Applying the values in the formula, we get

= (18x2-3y)(es) + (-3x+4y)(-sin s)

= (18e2s-3cos s)(es) + (-3es+4cos s)(-sin s)

= 18e3s-3escos s +3essin s - 4cos s sin s

At s = 0

= 18e3(0)-3e(0)cos 0 +3e(0)sin 0 - 4cos 0 sin 0

= 18-3

= 15

Problem 2 :

If

z(x,y) = x tan-1(xy), x = t2, y = set, s,t ∈ ℝ.

Find ∂z/∂s and ∂z/∂t at s = t = 1

Solution :

z(x, y) = x tan-1(xy)

∂z/∂s = (∂z/∂x)(∂x/∂s) + (∂z/∂y)(∂y/∂s) ---(1)

∂z/∂x = x[1/(1+(xy)2)] y+ tan-1(xy)(1)

∂z/∂x = [xy/(1+(xy)2)] + tan-1(xy)

∂x/∂s = 0

∂z/∂y = x[1/(1+(xy)2)] x+ tan-1(xy)(1)

∂z/∂y = [x2/(1+(xy)2)] + tan-1(xy) (0)

∂z/∂y =  [x2/(1+(xy)2)]

 y = set

∂y/∂s = et

By applying the values in the formula, we get

=[x2/(1+(xy)2)] + tan-1(xy)(0) + [x2/(1+(xy)2)] et

= [t4/(1+(t2set)2)] et

Applying the values of s and t, we get

= e/(1+e2)

∂z/∂s = (∂z/∂x)(∂x/∂s) + (∂z/∂y)(∂y/∂s) ---(2)

∂z/∂x = x[1/(1+(xy)2)] y+ tan-1(xy)(1)

∂z/∂x = [xy/(1+(xy)2)] + tan-1(xy)

∂x/∂t = 2t

∂z/∂y = x[1/(1+(xy)2)] x+ tan-1(xy)(1)

∂z/∂y = [x2/(1+(xy)2)] + tan-1(xy) (0)

∂z/∂y =  [x2/(1+(xy)2)]

∂y/∂t = set

∂z/∂t = {[xy/(1+(xy)2)] + tan-1(xy)}2t + [x2/(1+(xy)2)]set

∂z/∂t = 2t{[t2(set)/(1+(t2(set))2)] + tan-1(t2set)}2t + [t4/(1+(etset)2)]set

When s = t = 1

∂z/∂t = 2e/(1+e2+ 2tan-1(e) + e/(1+e2)

∂z/∂t = 3e/(1+e2+ 2tan-1(e)

Problem 3 :

Let

U(x, y) = ex sin y

where x = st2, y = s2t s,t ∈ ℝ. Find ∂U/∂s, ∂U/∂t and evaluate them at s = t = 1.

Solution :

∂U/∂s = (∂U/∂x)(∂x/∂s) + (∂U/∂y)(∂y/∂s) ----(1)

∂U/∂x = ex sin y

∂x/∂s = t2

∂U/∂y = ex cos y

∂y/∂s = 2st

Applying the values in the formula above.

= (ex sin y)t2 ex cos y(2st)

ex[(t2sin y) + 2st cos y]

= tex[t sin y + 2s cos y]

Applying the values of x and y, we get

= test^2[t sin (s2t) + 2s cos (s2t)]

Applying the value of s and t, we get

= test^2[sin (1) + 2cos (1)]

∂U/∂t = (∂U/∂x)(∂x/∂t) + (∂U/∂y)(∂y/∂t) ----(2)

x = st2, y = s2t

∂x/∂t = 2st, ∂y/∂t = s2

Applying the values in the formula above.

= (ex sin y)(2st) + ex cos y(s2)

= ex [2st sin y + cos y(s2)]

est^2 [2st sin (s2t) + cos (s2t)(s2)]

By applying the values of s and t, we get

[2sin (1) + cos (1)]

Problem 4 :

Let

z(x, y) = x3-3x2y3

where x = set, y = se-t, s, t  ∈ ℝ. Find ∂z/∂s, ∂z/∂t.

Solution :

∂z/∂s = (∂z/∂x) (∂x/∂s) + (∂z/∂y) (∂y/∂s) ---(1)

∂z/∂x = 3x2-3(2x)y3

∂z/∂x = 3x2-6xy3

∂x/∂s = et

∂z/∂y = 0-3x2(3y2)

∂z/∂y = -9x2y2

∂y/∂s = e-t

= (3x2-6xy3)et -9x2y2e-t

= 3etx2-6xy3e-9x2y2e-t

Applying the values of x and y

3et(set)2-6(set)(se-t)3e-9(set)2(se-t)2e-t

3s2e3t-6s4e-t -9s4e-t

3s2et[e2t-2s2e-2t -3s2]

∂z/∂t = (∂z/∂x) (∂x/∂t) + (∂z/∂y) (∂y/∂t) ---(1)

∂x/∂t = set and ∂y/∂t = -se-t

= (3x2-6xy3)set -9x2y2(-se-t)

Applying the values of x and y

= (3s2e2t-6s2)(set) - 9(s4)(-se-t)

= 3s3e3t-6s4e-2t + 9s5e-t

3s3[e3t-2se-2t +9s2e-t]

Problem 5 :

W(x, y, z) = xy + yz + zx, x = u − v, y = uv, z = u + v, u,v  ∈ ℝ. Find ∂W/∂u, ∂W/∂v and evaluate them at (1/2, 1).

Solution :

∂W/∂u = (∂W/∂x) (∂x/∂u) + (∂W/∂y) (∂y/∂u) + (∂W/∂z) (∂z/∂u)

∂W/∂x = y+0+z

∂W/∂x = y+z

∂x/∂u = 1

∂W/∂y = x+z+0

∂W/∂y = x+z

∂y/∂u = v

∂W/∂z = 0+y+x

∂W/∂z = y+x

∂z/∂u = 1

∂W/∂u = (y+z)(1) + (x+z)(v) + (y+x)(1)

∂W/∂u = 2y+x+z + (x+z)(v)

Applying the values of x = u − v, y = uv, z = u + v

∂W/∂u = 2uv+u-v+u+v + (u-v+u+v)(v)

∂W/∂u = 2uv+2u+uv-v2+uv+v2

∂W/∂u = 4uv+2u

∂W/∂u at (1/2, 1) = 4(1/2)(1)+2(1/2)

= 2 + 1

= 3

∂W/∂v = (∂W/∂x) (∂x/∂v) + (∂W/∂y) (∂y/∂v) + (∂W/∂z) (∂z/∂v)

∂W/∂x = y+0+z

∂W/∂x = y+z

∂x/∂v = -1

∂W/∂y = x+z+0

∂W/∂y = x+z

∂y/∂v = u

∂W/∂z = 0+y+x

∂W/∂z = y+x

∂z/∂v = 1

∂W/∂u = (y+z)(-1) + (x+z)(u) + (y+x)(1)

Simplifying and applying the values of 

x = u − v, y = uv, z = u + v

∂W/∂u = -y-z + (u-v+u+v)(u) + (uv+u-v)(1)

∂W/∂u = -uv-u-v + 2u2+ uv+u-v

∂W/∂u = -2v + 2u2

∂W/∂u at (1/2, 1) = -2(1) + 2(1/2)2

= -2 + 1/2

=  -3/2

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