Homework Soursefxt x2t t3x1 Find dfdt My guess is that dfdt partialdfdt
f(x,t) = x^2t + t^3(x-1)
Find df/dt.
My guess is that df/dt = partial_df/dt + partial_df/dx*dx/dt ...not sure how to calculate dx/dt
Find df/dt.
My guess is that df/dt = partial_df/dt + partial_df/dx*dx/dt ...not sure how to calculate dx/dt
Solution
The Chain Rule Review Of The Chain Rule For One Variable Recall that if y = f(x) and x = x(t) then dy dy dx = dt dx dt Example Suppose that z = f(x,y) = x2 + 2x - xy + y2 and x(t) = t2 +1 y(t) = t3 - t2 Then what is dz/dt for t = 2? Solution: f can be written as [x(t)]2 + 2[x(t)] - x(t)y(t) + [y(t)]2 Hence the derivative is 2x(t)[x\'(t)] + 2(2t) - [x(t)y\'(t) + y(t)x\'(t)] + 2y(t)y\'(t) = 2x(t)[2t] + 2(2t) - [x(t)(3t2 - 2t) + y(t)(2t)] + 2y(t)(3t2 - 2t) Now since t = 2, x(2) = 5 and y(2) = 4 We can substitute in to get 2(5)[2(2)] + 2(2)(2) - [(5)(3(2)2- 2(2)) + (4)2(2))] + 2(4)(3(2)2- 2(2)) = 56 Alternatively we use the chain rule: A Chain Rule With this chain rule the derivative becomes (2x + 2 - y)(2t) + (-x + 2y)(3t2 - 2t) When t = 2, x(2) = 5 and y(2) = 8 - 4 = 4 hence df / dt = [2(5) + 2 - 4](2)(2) + [-5 + 2(4)] [3(4) - 2(2)] = 56 Exercise: Let f(x,y) = 2x -3xy and x(q) = 2 cos q and y(q) = 2 sin q
