Question

if u=XY+yz+zx,X=1/t,y=e^tz=e^-t find du/dt​

Answer 1

If u = xy + yz + zx, where x=1/t, y=e^t, z=e^-t, then

du/dt = (1/t) (e^t - e^(-t)) - (1/t^2)(e^t + e^-t)

Given:
u = xy + yz + zx, where x=1/t, y=e^t, z=e^-t

To find:

du/dt​

Solution:

u = xy + yz + zx

=> du/dt = d(xy + yz + zx)/dt

=> du/dt = d(xy)/dt + d(yz)/dt + d(zx)/dt

(Using uv rule, d(uv)/dx = u d(v)/dx + v d(u)/dx)

=> du/dt = x d(y)/dt + y d(x)/dt + y d(z)/dt + z d(y)/dt + z d(x)/dt + x d(z)/dt

Substituting the values of x, y and z in this expression, we get

=> du/dt = (1/t) d(e^t)/dt + e^t d(1/t)/dt + e^t d(e^-t)/dt + e^-t d(e^t)/dt + e^-t                     d(1/t)/dt + (1/t) d(e^-t)/dt

(We know

d(1/x)/dx = -1/x^2; d(e^x)/dx = e^x)

=> du/dt = (1/t) e^t + e^t (-1/t^2) + e^t (-e^-t) + e^-t e^t + e^-t (-1/t^2) + (1/t)(-e^-t)

=> du/dt = (1/t) e^t + e^t (-1/t^2) - 1 + 1 + e^-t (-1/t^2) + (1/t)(-e^-t)

=> du/dt = (1/t) (e^t - e^(-t)) - (1/t^2)(e^t + e^-t)

Hence,

du/dt = (1/t) (e^t - e^(-t)) - (1/t^2)(e^t + e^-t)

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