Daisy
Lv 4
Daisy asked in Science & MathematicsMathematics · 7 years ago

derivative help please :) 10 points for the best answer!!?

find the derivative of f(k)= sqrt [k+ (k-c)^2] with respect to k.

3 Answers

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  • Julia
    Lv 5
    7 years ago
    Favorite Answer

    ƒ(k) = √[k + (k − c)^2], assuming c is a constant, and k is the only variable.

    ƒ(k) = [k + (k − c)^2]^(1/2)

    Let:

    u = k + (k − c)^2

    v = 1/2

    ƒ(k) = u^v, by chain rule, we get:

    ƒ'(k) = u' ⋅ v ⋅ u^(v - 1)

    ƒ'(k) = [1 + 2(k − c)] ⋅ 1/2 ⋅ [k + (k − c)^2]^(-1/2)

    ƒ'(k) = (1 + 2k − 2c) / (2√[k + (k − c)^2])

    ƒ'(k) = (1/2 + k − c) / (√[k + (k − c)^2])

  • oldman
    Lv 5
    7 years ago

    f(k)= sqrt [k+ (k-c)^2] ===> d/dx (x-c)² = 2(x-c)(1) = 2(x-c)

    f(k)= sqrt [k+ (k-c)^2] ===>d/dx (x + x²)^(1/2)= (1/2) [ x + x²)( 1 + 2x)

    f(k)= sqrt [k+ (k-c)^2] ===>d/dk (1/2) [ k + ( k-c)² ] [ (1+2(k-c)]

    d/dk (1/2) [ (1+2(k-c)] [ k + ( k-c)² ] = [(1/2) + (k-c)] [ k + ( k-c)² ]

  • 7 years ago

    1 - 2c + 2k

    -------------------------------

    2√(c² - 2ck + k² + k))

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