Question

f(x, y) = x^2 + y^3 ; X = t^2 + t^3; y = t^3 + t^9 Find ^df⁄dt at t=1.

(a) 0

(b) 1

(c) -1

(d) 164

I had been asked this question in an interview for internship.

Query is from Partial Differentiation topic in portion Partial Differentiation of Engineering Mathematics

Answer 1

Correct option is (d) 164

For explanation: Using chain rule we have

\(\frac{df}{dt}=f_x.\frac{dx}{dt}+f_y.\frac{dy}{dt}\)

=(2x).(2t + 3t^2) + (3y^2).(3t^2 + 9t^8)

Put t = 1; we have x = 2; y = 2

=4.(5) + 12.(12) = 164.