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chetibaby
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Tyro
Asked: 2021-01-10T23:13:17+00:00 In:

Differentiate y=(2x-1)(x+4)^{2} wrt x

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Differentiate y=(2x-1)(x+4)^{2} wrt x

differentiationproduct rule

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2 Answers

  2. Sengemegz
    Moderator

    y=(2x-1)(x+4)^{2}
    Let u=2x-1 and v=(x+4)^{2}
    \dfrac{du}{dx}=2, \dfrac{dv}{dx}=2(x+4)
    \Rightarrow \dfrac{dy}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}
    \Rightarrow \dfrac{dy}{dx}=((x+4)^{2}\times 2) + ((2x-1)\times (2x+8))
    \Rightarrow \dfrac{dy}{dx}=2(x+4)^{2}+4x^{2}+16x-2x-8
    \Rightarrow \dfrac{dy}{dx}=2(x^{2}+8x+16)+4x^{2}+14x-8
    \Rightarrow \dfrac{dy}{dx}=2x^{2}+16x+32+4x^{2}+14x-8
    \therefore \dfrac{dy}{dx}=6x^{2}+30x+24

  4. acedstud
    Best Answer

    To differentiate the function y = (2x - 1)(x + 4)^2 with respect to x, you can use the product rule. The product rule states that if you have a function of the form u(x)v(x), the derivative is given by:

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

    In this case, let u(x) = 2x - 1 and v(x) = (x + 4)^2.

    Now, we’ll find their derivatives:

    u'(x) = \dfrac{d}{dx} (2x - 1) = 2 (the derivative of 2x is 2, and the derivative of a constant, -1, is 0).

    To find v'(x), you can use the chain rule. The chain rule states that if you have a composite function, such as v\left(w(x)\right), the derivative of v\left(w(x)\right) with respect to x is given by:

    \dfrac{d}{dx}\left[v\left(w(x)\right)\right] = v'\left(w(x)\right) * w'(x).

    Let w(x) = x + 4, so v(x) = w(x)^2. Now, we can find v'(x):

    v'(x) = \dfrac{d}{dx} \left[w(x)^2\right] = 2w(x)w'(x).

    w'(x) = \dfrac{d}{dx} (x + 4) = 1 (the derivative of x is 1, and the derivative of a constant, 4, is 0).

    So, v'(x) = 2(x + 4) * 1 = 2(x + 4).

    Now, apply the product rule:

    \dfrac{d}{dx}\left[(2x - 1)(x + 4)^2\right] = u(x)v'(x) + u'(x)v(x) = (2x - 1) * 2(x + 4) + 2 * (x + 4)^2.

    Now, simplify this expression:

    (4x - 2)(x + 4) + 2(x + 4)^2.

    You can further simplify this expression if needed, but this is the derivative of y = (2x - 1)(x + 4)^2 with respect to x:

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

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