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chetibaby
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Tyro
Asked: 2021-01-10T23:18:10+00:00 In:

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

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

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2 Answers

  1. acedstud
    Best Answer

    To differentiate the function y = x^2(x^3 - 1)^3 with respect to x, we 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:

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

    In this case, let u(x) = x^2 and v(x) = (x^3 - 1)^3. We need to find their derivatives:

    1. u'(x) = \frac{d}{dx}(x^2) = 2x (using the power rule).

    2. We can find v'(x) using 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:

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

    To find v'(x), let w(x) = x^3 - 1, so v(x) = w(x)^3:

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

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

        \[\frac{d}{dx}[w(x)] &= \frac{d}{dx}(x^3 - 1) \\ &= 3x^2\]

    Now, we can compute v'(x):

        \[v'(x) &= 3w(x)^2 \cdot 3x^2 \\ &= 9x^2(x^3 - 1)^2\]

    Now, apply the product rule to find \frac{d}{dx}[x^2(x^3 - 1)^3]:

        \[\frac{d}{dx}[x^2(x^3 - 1)^3] &= u(x)v'(x) + u'(x)v(x) \\ &= x^2 \cdot 9x^2(x^3 - 1)^2 + 2x \cdot (x^3 - 1)^3\]

    Now, simplify this expression:

        \[\frac{d}{dx}[x^2(x^3 - 1)^3] &= 9x^4(x^3 - 1)^2 + 2x(x^3 - 1)^3\]

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

        \[\frac{d}{dx}[x^2(x^3 - 1)^3] &= 9x^4(x^3 - 1)^2 + 2x(x^3 - 1)^3\]

  3. Sengemegz
    Moderator
    This answer was edited.

    y=x^{2}(x^{3}-1)^{3}
    Let u=x^{2} and v=(x^{3}-1)^{3}
    \dfrac{du}{dx}=2x, \dfrac{dv}{dx}=6x^{2}(x^{3}-1)^{2}
    \Rightarrow \dfrac{dy}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}
    \Rightarrow \dfrac{dy}{dx}=((x^{3}-1)^{3}\times 2x)+(x^{2}\times 6x^{2}(x^{3}-1)^{2})
    \Rightarrow \dfrac{dy}{dx}=2x(x^{3}-1)^{3}+6x^{4}(x^{3}-1)^{2}
    \Rightarrow \dfrac{dy}{dx}=(x^{3}-1)^2(2x(x^{3}-1)+6x^{4})
    \Rightarrow \dfrac{dy}{dx}=(x^{3}-1)^2(2x^{4}-2x+6x^{4})
    \therefore \dfrac{dy}{dx}=(x^{3}-1)^2(8x^{4}-2x)

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