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COLLEGE CALCULUS. HELP!

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Find the derivative of the function.

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

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

  • COLLEGE CALCULUS. HELP! - ,

    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

  • COLLEGE CALCULUS. HELP! - ,

    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]

  • COLLEGE CALCULUS. HELP! - ,

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

  • COLLEGE CALCULUS. HELP! - ,

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

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