Question

Find the derivative of the function.

1. F(x)= (1+2x+x^3)^1/4

2. f(t)= (1+tant)^1/3

Answer 1

1.
1 turns to 0
2x turns to 2
x^3 turns to 3x^2 (bring 3 up front & subtract 1 from exponent)
^1/4 goes up front and subtract 1 from that =-3/4

Final answer would be

1/4(3x^2+2)^-3/4

2. Use same procedure
1/3 (sec^2t) ^ -2/3

Answer 2

d/dx (u^n) = n u^(n-1) du/dx
Not quite Will.
I get
(1/4)(x^3+2x+2)^-3/4 [3x^2+2]

Answer 3

f(t)= (1+tant)^1/3
f'(t) = (1/3)(1+tan t)^-(2/3) [ sec^2 t]

Answer 4

Damon is right, I forgot to apply the chain rule.