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 = e^t and y = sin t

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

Solution :

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

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

du/dt = 2e^txy+3e^ty⁴ + x²cos t+12xy³cos t

x = e^t and y = sin t

du/dt = 2e^te^tsint+3e^tsin⁴t + (e^t)²cos t+12e^tsin³tcos t

du/dt = e^t(2e^tsint+3sin⁴t + e^tcos t+12sin³tcos t)

When t = 0

du/dt = e⁰(2e⁰sin(0)+3sin⁴0 + e^tcos 0+12sin³0cos 0)

du/dt = 1

Problem 2 :

If

u(x, y, z) = xy²z³, x = sint, y = cost, z = 1+e^2t

find du/dt

Solution :

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

du/dt = y²z³cost + 2xyz³(-sint) + 3xy²z²(2e^2t)

Applying x = sint t, y = cost and z = 1+e^2t

du/dt = cos²t(1+e^2t)³cost + 2sintcost (1+e^2t)³(-sint) + 3sint cos²t (1+e^2t)²(2e^2t)

= (1+e^2t)²[(1+e^2t) cos³t - sin2tsint(1+e^2t) + 6e^2tsint cos²t]

Problem 3 :

If

w(x, y, z) = x²+y²+z², x = e^t, y = e^t sint and z = e^tcost

find dw/dt.

Solution :

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

∂w/∂x = 2x

dx/dt = e^t

By applying the values in (1), we get

= 2xe^t + 2ye^t[cost+sint] + 2ze^t[cost-sint]

= 2e^t[x+y(cost+sint)+z(cost-sint)]

= 2e^t[e^t+e^t sint(cost+sint)+e^tcost(cost-sint)]

= 2e^2t[1+sintcost+sin²t+cos²t-sintcost]

= 2e^2t(1+1)

= 4 e^2t

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

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]

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