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April 18, 2014

April 18, 2014

Posted by **cynthia** on Sunday, February 26, 2012 at 1:43am.

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

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

- COLLEGE CALCULUS. HELP! -
**Will**, Sunday, February 26, 2012 at 2:01am1.

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

- COLLEGE CALCULUS. HELP! -
**Damon**, Sunday, February 26, 2012 at 3:30amd/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]

- COLLEGE CALCULUS. HELP! -
**Damon**, Sunday, February 26, 2012 at 3:32amf(t)= (1+tant)^1/3

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

- COLLEGE CALCULUS. HELP! -
**Will**, Sunday, February 26, 2012 at 3:48amDamon is right, I forgot to apply the chain rule.

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