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Help with derivatives question?

What is f'(x) if f(x) = cos x^2

8 Answers

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  • 2 months ago

    Since no one actually explained it, I'll do that, here.

    Presuming that is:

    f(x) = cos(x²)

    We can set up the chain rule:

    f = cos(u) and u = x²

    Get the derivative of both:

    df/du = -sin(u) and du/dx = 2x

    Now we can apply the chain rule:

    df/dx = df/du * du/dx

    df/dx = -sin(u) * 2x

    df/dx = -2x sin(u)

    df/dx = -2x sin(x²)

    Back into function notation:

    f'(x) = -2x sin(x²)

  • 2 months ago

    f(x)=cos(x^2)

    =>

    f '(x)=-sin(x^2)(2x)

    =>

    f '(x)=-2xsin(x^2)

  • 2 months ago

    f(x) = cos(x^2)

    f'(x) = - 2xsin(x^2) answer//

  • 2 months ago

    d/dx(cos(x^2)) = -2 x sin(x^2)

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  • 2 months ago

    f'(x) = -2x sin x^2

  • 2 months ago

    First way:

    f(x) = cos²(x)

    Recall the derivative of u^(n) is: n * u' * n^(n - 1) → where in your case:

    u = cos(x) → u' = - sin(x)

    n = 2

    f'(x) = n * u' * n^(n - 1)

    f'(x) = 2 * [- sin(x)] * [cos(x)]^(2 - 1)

    f'(x) = - 2.sin(x).cos(x)

    Second way:

    f(x) = cos(x) * cos(u)

    Recall the derivative of u^(n) is: (u'.v) + (v'.u) → where in your case:

    u = cos(x) → u' = - sin(x)

    v = cos(x) → v' = - sin(x)

    f'(x) = (u'.v) + (v'.u)

    f'(x) = [- sin(x) * cos(x)] + [- sin(x) * cos(x)]

    f'(x) = - sin(x).cos(x) - sin(x).cos(x)

    f'(x) = - 2.sin(x).cos(x)

  • 2 months ago

    I think you mean f(x) = cos²x

    Using the chain rule we can show that:

    f '(x) = -2cosxsinx

    Now, 2cosxsinx => sin2x

    so, f '(x) = -sin2x

    :)>

  • Bryce
    Lv 7
    2 months ago

    f'(x)= -2cosx*sinx

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