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chetibaby
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Tyro
Asked: 2021-01-10T23:21:53+00:00 In:

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

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

differentiationproduct rule

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2 Answers

  2. acedstud
    Best Answer
    This answer was edited.

    To differentiate the function y = \sqrt{x}(x+3)^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:

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

    In this case, let u(x) = \sqrt{x} and v(x) = (x+3)^2. We need to find their derivatives:

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

    2. v'(x) = \frac{d}{dx}((x+3)^2) = 2(x+3) (using the chain rule).

    Now, apply the product rule:

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

    Now, simplify this expression:

        \[2\sqrt{x}(x + 3) + \frac{1}{2\sqrt{x}}(x+3)^2\]

    So, the is the derivative of y = \sqrt{x}(x+3)^2 with respect to x:

        \[2\sqrt{x}(x + 3) + \frac{1}{2\sqrt{x}}(x+3)^2\]

  4. Sengemegz
    Moderator

    y=sqrt{x}(x+3)^{2}
    Let u=x^{\frac{1}{2}} and v=(x+3)^{2}
    \dfrac{du}{dx}=\dfrac{1}{2}x^{-\frac{1}{2}}, \dfrac{dv}{dx}=2(x+3)
    \Rightarrow \dfrac{dy}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}
    \Rightarrow \dfrac{dy}{dx}=((x+3)^{2}\times \dfrac{1}{2}x^{-\frac{1}{2}})+(x^{\frac{1}{2}}\times (2x+6))
    \Rightarrow \dfrac{dy}{dx}=\dfrac{(x+3)^{2}}{2\sqrt{x}}+\sqrt{x}(2x+6)
    \Rightarrow \dfrac{dy}{dx}=\dfrac{(x+3)^{2}+2\sqrt{x}\times \sqrt{x}(2x+6)}{2\sqrt{x}}
    \Rightarrow \dfrac{dy}{dx}=\dfrac{(x+3)^{2}+2x(2x+6)}{2\sqrt{x}}
    \Rightarrow \dfrac{dy}{dx}=\dfrac{x^{2}+6x+9+4x^{2}+12x}{2\sqrt{x}}
    \Rightarrow \dfrac{dy}{dx}=\dfrac{5x^{2}+18x+9}{2\sqrt{x}}
    \Rightarrow \dfrac{dy}{dx}=\dfrac{(x+3)(x+15)}{2\sqrt{x}}

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