CHAIN RULE IN PARTIAL DERIVATIVES

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

u(x, y) = x2y+3xy4, x = et and y = sin t

find du/dt and evaluate it at t = 0.

Solution :

du/dt = (∂u/∂x) (dx/dt) + (∂u/∂y) (dy/dt)

∂u/∂x = 2xy+3(1)y4

∂u/∂x = 2xy+3y4

dx/dt = et

∂u/∂y = x2(1)+3x(4y3)

∂u/∂y = x2+12xy3

dy/dt = cos t

du/dt = (2xy+3y4)e+ (x2+12xy3)cos t

du/dt = 2etxy+3ety4 + x2cos t+12xy3cos t

x = et and y = sin t

du/dt = 2etetsint+3etsin4t + (et)2cos t+12etsin3tcos t

du/dt et(2etsint+3sin4t + etcos t+12sin3tcos t)

When t = 0

du/dt e0(2e0sin(0)+3sin40 + etcos 0+12sin30cos 0)

du/dt = 1

Problem 2 :

If

u(x, y, z) = xy2z3, x = sint, y = cost, z = 1+e2t

find du/dt

Solution :

du/dt = (∂u/∂x) (dx/dt) + (∂u/∂y) (dy/dt) + (∂u/∂z) (dz/dt)

∂u/∂x = (1)y2z3

∂u/∂x = y2z3

dx/dt = cost

∂u/∂y = x(2y)z3

∂u/∂y = 2xyz3

dy/dt = -sint

∂u/∂z = xy2(3z2)

∂u/∂z = 3xy2z2

dz/dt = 0+e2t(2)

dz/dt = 2e2t

du/dt = y2z3cost + 2xyz3(-sint) + 3xy2z2(2e2t)

Applying x = sint t, y = cost and z = 1+e2t

du/dt = cos2t(1+e2t)3cost + 2sintcost (1+e2t)3(-sint) + 3sint cos2t (1+e2t)2(2e2t)

= (1+e2t)2[(1+e2t) cos3t - sin2tsint(1+e2t) + 6e2tsint cos2t]

Problem 3 :

If

w(x, y, z) = x2+y2+z2, x = et, y = et sint and z = etcost

find dw/dt.

Solution :

dw/dt = (∂w/∂x) (dx/dt) + (∂w/∂y) (dy/dt) + (∂w/∂z) (dz/dt)  ----(1)

∂w/∂x = 2x

dx/dt = et

∂w/∂y = 2y

dy/dt = et(cost)+sint(et)

dy/dt et[cost+sint]

∂w/∂z = 2y

dz/dt = et(-sint)+cost(et)

dz/dt et[cost-sint]

By applying the values in (1), we get

2xet + 2yet[cost+sint] + 2zet[cost-sint]

= 2et[x+y(cost+sint)+z(cost-sint)]

= 2et[et+et sint(cost+sint)+etcost(cost-sint)]

= 2e2t[1+sintcost+sin2t+cos2t-sintcost]

= 2e2t(1+1)

= 4 e2t

Problem 4 :

Let

U(x, y, z) = xyz, x = e-t, y = e-tcost, z = sint, t ℝ.

Find dU/dt.

Solution :

dU/dt = (∂U/∂x) (dx/dt) + (∂U/∂y) (dy/dt) + (∂U/∂z) (dz/dt) ----(1)

∂U/∂x = yz

dx/dt = -e-t

∂U/∂y = xz

y = e-tcost

dy/dt

e-t(-sint)+cost(-e-t)

= e-t(-sint-cost)

∂U/∂z = xy

z = sint

dz/dt = cost

Applying the values in (1), we get

= yz(-e-t) + xze-t(-sint-cost) + xy (cost)

Applying the values of x = e-t, y = e-tcost, z = sint

Part 1 :

yz(-e-t)  ==> e-tcost sint (-e-t) ==>  -e-2tsint cost 

Part 2 :

xze-t(-sint-cost) ==> e-tsinte-t(-sint-cost)

= e-2t sint(-sint-cost)

= e-2t sint(-sint-cost)

Part 3 :

xy (cost) = e-te-tcost (cost)

= e-2t cos2t

Adding part 1, 2 and 3, we get

= e-2t[-sint cost-sin2t-sint cost+cos2t]

= e-2t[cos2t-sin2t - 2sint cost]

= e-2t[cos2t - sin2t]

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Trigonometry Word Problems Worksheet with Answers

    Jan 17, 22 10:45 AM

    Trigonometry Word Problems Worksheet with Answers

    Read More

  2. Trigonometry Word Problems with Solutions

    Jan 17, 22 10:41 AM

    Trigonometry Word Problems with Solutions

    Read More

  3. Writing Numbers in Words Worksheet

    Jan 16, 22 11:56 PM

    Writing Numbers in Words Worksheet

    Read More