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chetibaby
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Tyro
Asked: 2021-01-10T23:05:52+00:00 In:

Differentiate y=x^{2}(2x-5)^{4} wrt x

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Differentiate y=x^{2}(2x-5)^{4} wrt x

differentiationproduct rule

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2 Answers

  2. acedstud
    Best Answer
    This answer was edited.

    To differentiate the function y = x^2(2x - 5)^4 with respect to x, you 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) = x^2 and v(x) = (2x - 5)^4.

    Now, we’ll find their derivatives:

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

    To find v'(x), we’ll need to 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) = 2x - 5, so v(x) = w(x)^4. Now, we can find v'(x):

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

    w'(x) = \dfrac{d}{dx} (2x - 5) = 2 (the derivative of 2x is 2, and the derivative of a constant, -5, is 0).

    So,

    v'(x) = 4(2x - 5)^3 * 2 = 8(2x - 5)^3.

    Now, we can apply the product rule:

    \dfrac{d}{dx}\left[x^2(2x - 5)^4\right] = u(x)v'(x) + u'(x)v(x) = x^2 * 8(2x - 5)^3 + 2x * (2x - 5)^4.

    Now, simplify this expression:

    8x^2(2x - 5)^3 + 2x(2x - 5)^4.

    So, the corrected derivative of y = x^2(2x - 5)^4 with respect to x is:

    y'(x) = 8x^2(2x - 5)^3 + 2x(2x - 5)^4.

  4. Sengemegz
    Moderator
    This answer was edited.

    y=x^{2}(2x-5)^{4}
    Let u=x^{2} then \dfrac{du}{dx}=2x
    and v=(2x-5)^{4} then \dfrac{dv}{dx}=8(2x-5)^{3}
    \dfrac{dy}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}
    \Rightarrow \dfrac{dy}{dx}=((2x-5)^{4}\times 2x)+(x^{2}\times 8(2x-5)^{3})
    \Rightarrow \dfrac{dy}{dx}=2x(2x-5)^{4}+8x^{2}(2x-5)^{3}
    \Rightarrow \dfrac{dy}{dx}=(2x-5)^{3}(2x(2x-5)+8x^{2})
    \Rightarrow \dfrac{dy}{dx}=(2x-5)^{3}(4x^{2}-10x+8x^{2})
    \therefore \dfrac{dy}{dx}=(2x-5)^{3}(12x^{2}-10x)

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