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chetibaby
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Tyro
Asked: 2021-01-10T23:04:05+00:00 In:

Differentiate y=(3x-2)(x^{2}+3) wrt x

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Differentiate y=(3x-2)(x^{2}+3) wrt x

differentiationproduct rule

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2 Answers

  2. Sengemegz
    Moderator
    This answer was edited.

    y=(3x-2)(x^{2}+3)
    Let u=3x-2 then \dfrac{du}{dx}=3
    and v=x^{2}+3 then \dfrac{dv}{dx}=2x
    \dfrac{dy}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}
    \Rightarrow \dfrac{dy}{dx}=((x^{2}+3)\times 3)+((3x-2)\times 2x)
    \Rightarrow \dfrac{dy}{dx}=3(x^{2}+3)+2x(3x-2)
    \Rightarrow \dfrac{dy}{dx}=3x^{2}+9+6x^{2}-4x
    \Rightarrow \dfrac{dy}{dx}=9x^{2}-4x+9

  4. acedstud
    Best Answer

    To differentiate the function y = (3x - 2)(x^2 + 3) with respect to x, you can use the product rule, which states that the derivative of a product of two functions is given by:

    \frac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x),

    where u(x) and v(x) are functions of x, and u'(x) and v'(x) are their respective derivatives with respect to x.

    In this case, let u(x) = 3x - 2 and v(x) = x^2 + 3.

    Now, let’s find their derivatives:

    u'(x) = \dfrac{d}{dx} (3x - 2) = 3, (the derivative of 3x is 3, and the derivative of a constant, -2, is 0).

    v'(x) = \dfrac{d}{dx} (x^2 + 3) = 2x, (use the power rule to find the derivative of x^2, which is 2x, and the derivative of a constant, 3, is 0).

    Now, apply the product rule:

    \dfrac{d}{dx}\left[(3x - 2)(x^2 + 3)\right] = (3x - 2)(2x) + (3)(x^2 + 3).

    Now, simplify this expression:

    (3x - 2)(2x) + 3(x^2 + 3) = 6x^2 - 4x + 3x^2 + 9.

    Combine like terms:

    6x^2 + 3x^2 - 4x + 9 = 9x^2 - 4x + 9.

    So, the derivative of y = (3x - 2)(x^2 + 3) with respect to x is:

    y'(x) = 9x^2 - 4x + 9.

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