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chetibaby
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Tyro
Asked: 2021-01-10T23:11:34+00:00 In:

Differentiate y=3x^{3}(x^{2}+4)^{2} wrt x

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Differentiate y=3x^{3}(x^{2}+4)^{2} wrt x

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2 Answers

  2. Sengemegz
    Moderator

    y=3x^{3}(x^{2}+4)^{2}
    Let u=3x^{3} and v=(x^{2}+4)^{2}
    \dfrac{du}{dx}=9x^{2}, \dfrac{dv}{dx}=4x(x^{2}+4)
    \Rightarrow \dfrac{dy}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}
    \Rightarrow \dfrac{dy}{dx}=((x^{2}+4)^{2}\times 9x^{2}) + (3x^{3}\times 4x^{3}+16x)
    \Rightarrow \dfrac{dy}{dx}=9x^{2}(x^{2}+4)^{2}+12x^{6}+48x^{4}
    \Rightarrow \dfrac{dy}{dx}=9x^{2}(x^{4}+8x^{2}+16)+12x^{6}+48x^{4}
    \Rightarrow \dfrac{dy}{dx}=9x^{6}+72x^{4}+144x^{2}+12x^{6}+48x^{4}
    \Rightarrow \dfrac{dy}{dx}=21x^{6}+120x^{4}+144x^{2}

  4. acedstud
    Best Answer

    To differentiate the function y = 3x^3(x^2 + 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) = 3x^3 and v(x) = (x^2 + 4)^2.

    Now, we’ll find their derivatives:

    u'(x) = \dfrac{d}{dx} (3x^3) = 9x^2 (using the power rule).

    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^2 + 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^2 + 4) = 2x (the derivative of x^2 is 2x, and the derivative of a constant, 4, is 0).

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

    Now, apply the product rule:

    \dfrac{d}{dx}\left[3x^3(x^2 + 4)^2\right] = u(x)v'(x) + u'(x)v(x) = 3x^3 * 4x(x^2 + 4) + 9x^2(x^2 + 4)^2.

    Now, simplify this expression:

    12x^4(x^2 + 4) + 9x^2(x^2 + 4)^2.

    So, the derivative of y = 3x^3(x^2 + 4)^2 with respect to x is:

    y'(x) = 12x^4(x^2 + 4) + 9x^2(x^2 + 4)^2.

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