Use the chain rule to find dz/dt. z = sin(x) cos(y), x = t , y = 5 / t

Solution:

The chain rule of differentiation is given by

dz / dt = (dz / dx)(dx / dt) + (dz / dy)(dy / dt)

Given, z = sin(x) cos(y)

dz/dx = d(sinx cosy) / dx

dz/dx = cosx cosy

dx/dt = 1 

dz/dy = sinx(-siny)

dy/dt = 1/5

dz/dx = cosx cosy (1) - sinx siny(1/5)

dz/dx = 1/5[5(cosx cosy) - sinx siny]

Put x=t and y=t/5

dz/dx = 1/5 [5(cos(t) cos(t/5) - sin(t) sin(t/5)] ------------------------ (1)

We know, cosA cosB = 1/2[cos(A + B) + cos(A - B)]

cos(t) cos(t/5) = 1/2[cos(t + (t / 5)) + cos(t - (t / 5))]

cos(t) cos(t/5) = 1/2 [cos(6t / 5) + cos(4t / 5)] --------------------------- (2)

We know, sinA sinB = 1/2 [cos(A - B) - cos(A + B)]

sin(t) sin(t/5) = 1/2 [cos(t - (t/5)) - cos(t + (t/5))]

sin(t) si(t/5) = 1/2 [cos(4t / 5) - cos(6t / 5)] -------------------------------- (3)

Substitute (2) and (3) in (1)

dz/dt = 1/5 [5(1/2[cos(6t / 5) + cos(4t / 5)]) - 1/2 [cos(4t / 5)-cos(6t / 5)]]

Taking out 1/2 in RHS,

dz/dt = 1/10[ 5 cos(6t / 5)+5 cost(4t / 5) - cos(4t / 5) + cos(6t / 5)]

dz/dt = 1/10 [6 cos(6t / 5) + 4 cos(4t / 5)]

dz/dt = 1/10 [6 cos(6t / 5)] + 1/10 [4 cos(4t / 5)]

dz/dt = 3/5 [cos(6t / 5)] + 2 / 5[cos(4t / 5)]

Therefore, dz/dt by chain rule is 3 / 5[cos(6t / 5)] + 2 / 5[cos(4t / 5)]

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Use the chain rule to find dz/dt. z = sin(x) cos(y), x = t , y = 5 / t

Summary:

The value of dz/dt when z = sin(x)cos(y) for x = t and y = t / 5 using chain rule is 3/5[cos(6t / 5)] + 2/5 [cos(4t / 5)]