COLLEGE CALCULUS. HELP!
posted by cynthia .
Find the derivative of the function.
1. F(x)= (1+2x+x^3)^1/4
2. f(t)= (1+tant)^1/3

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 
d/dx (u^n) = n u^(n1) du/dx
Not quite Will.
I get
(1/4)(x^3+2x+2)^3/4 [3x^2+2] 
f(t)= (1+tant)^1/3
f'(t) = (1/3)(1+tan t)^(2/3) [ sec^2 t] 
Damon is right, I forgot to apply the chain rule.
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