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chetibaby
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Tyro
Asked: 2021-01-10T23:16:38+00:00 In:

Differentiate y=x(x^{2}-1)^{3} wrt x

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Differentiate y=x(x^{2}-1)^{3} wrt x

differentiationproduct rule

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2 Answers

  2. acedstud
    Best Answer
    This answer was edited.

    To differentiate the function y = x(x^2 - 1)^3 with respect to x, you can apply both the product rule and the chain rule.

    Using the product rule, which states that \dfrac{d}{dx}\left[u(x)v(x)\right] = u(x)v'(x) + u'(x)v(x), we have:

    Let u(x) = x and v(x) = (x^2 - 1)^3.

    We know that u'(x) = 1 (the derivative of x is 1).

    Now, we need to find v'(x), which involves applying the chain rule because v(x) is a composite function. 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 - 1, so v(x) = w(x)^3.

    Now, we can find v'(x):

        \[v'(x) &= \dfrac{d}{dx}[w(x)^3] \\ &= 3w(x)^2 \cdot \dfrac{d}{dx}[w(x)]\]

    Now, we find \dfrac{d}{dx}[w(x)]:

        \[\dfrac{d}{dx}[w(x)] &= \dfrac{d}{dx}[x^2 - 1] \\ &= 2x\]

    Now, we can compute v'(x):

        \[v'(x) &= 3w(x)^2 \cdot 2x \\ &= 6x(x^2 - 1)^2\]

    Now, apply the product rule:

        \[\dfrac{d}{dx}[x(x^2 - 1)^3] &= u(x)v'(x) + u'(x)v(x) \\ &= x \cdot 6x(x^2 - 1)^2 + 1 \cdot (x^2 - 1)^3\]

    Now, simplify this expression:

        \[\dfrac{d}{dx}[x(x^2 - 1)^3] &= 6x^2(x^2 - 1)^2 + (x^2 - 1)^3\]

    So, the correct derivative of y = x(x^2 - 1)^3 with respect to x is:

        \[\dfrac{d}{dx}[x(x^2 - 1)^3] &= 6x^2(x^2 - 1)^2 + (x^2 - 1)^3\]

  4. Sengemegz
    Moderator

    y=x(x^{2}-1)^{3}
    Let u=x and v=(x^{2}-1)^{3}
    \dfrac{du}{dx}=1, \dfrac{dv}{dx}=6x(x^{2}-1)^{2}
    \Rightarrow \dfrac{dy}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}
    \Rightarrow \dfrac{dy}{dx}=((x^{2}-1)^{3}\times 1) + (x\times 6x(x^{2}-1)^{2})
    \Rightarrow \dfrac{dy}{dx}=(x^{2}-1)^{3}+6x^{2}(x^{2}-1)^{2}
    \Rightarrow \dfrac{dy}{dx}=(x^{2}-1)^{2}(x^{2}-1+6x^{2})
    \therefore \dfrac{dy}{dx}=(x^{2}-1)^{2}(7x^{2}-1)

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