Question

Let f(x, y) = x²y + 3xy4, where x = Sin2t, y = Cos2t. Find the value of df/dt at t = 0. [Ans: 6]​

Answer 1

Step-by-step explanation:

df/dt =df/dx*dx/dt + df/dy*dy/dt

=(2xy+3y4)*2cos2t + (x2+12xy3)*-2sin2t

At t=0, second term becomes 0

x=0 and y=1

so the first term (2xy+3y4)*2cos2t reduces to 3*2=6