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chetibaby
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Tyro
Asked: 2021-01-10T23:23:25+00:00 In:

Differentiate y=x\sqrt{x+1} wrt x

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Differentiate y=x\sqrt{x+1} wrt x

differentiationproduct rule

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2 Answers

  2. Sengemegz
    Moderator
    This answer was edited.

    y=x\sqrt{x+1}

    Let u=x and v=(x+1)^{\frac{1}{2}}

    \dfrac{du}{dx}=1, \dfrac{dv}{dx}=\dfrac{1}{2}(x+1)^{-\frac{1}{2}}

    \Rightarrow \dfrac{dy}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}

    \Rightarrow \dfrac{dy}{dx}=((x+1)^{\frac{1}{2}}\times 1)+(x\times \dfrac{1}{2}(x+1)^{-\frac{1}{2}})

    \Rightarrow \dfrac{dy}{dx}=(x+1)^{\frac{1}{2}}+\dfrac{x}{2(x+1)^{\frac{1}{2}}}

    \Rightarrow \dfrac{dy}{dx}=\dfrac{2(x+1)^{\frac{1}{2}}\times (x+1)^{\frac{1}{2}}+x}{2(x+1)^{\frac{1}{2}}}

    \Rightarrow \dfrac{dy}{dx}=\dfrac{2(x+1)+x}{2(x+1)^{\frac{1}{2}}}

    \therefore \dfrac{dy}{dx}=\dfrac{3x+2}{2\sqrt{x+1}}

  4. acedstud
    Best Answer

    To differentiate the function y = x\sqrt{x+1} 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) = x and v(x) = \sqrt{x+1}. We need to find their derivatives:

    1. u'(x) = \frac{d}{dx}(x) = 1 (the derivative of x is 1).

    2. To find v'(x), you can use the chain rule. Let w(x) = x+1, so v(x) = \sqrt{w(x)}. Now, we can find v'(x):

        \[v'(x) &= \frac{d}{dx}[\sqrt{w(x)}] \\ &= \frac{1}{2\sqrt{w(x)}}\frac{d}{dx}[w(x)]\]

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

        \[\frac{d}{dx}[w(x)] &= \frac{d}{dx}(x+1) \\ &= 1\]

    Now, we can compute v'(x):

        \[v'(x) &= \frac{1}{2\sqrt{w(x)}}\cdot 1 \\ &= \frac{1}{2\sqrt{x+1}}\]

    Now, apply the product rule to find \frac{d}{dx}[x\sqrt{x+1}]:

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

    Now, simplify this expression:

        \[\frac{d}{dx}[x\sqrt{x+1}] &= \frac{x}{2\sqrt{x+1}} + \sqrt{x+1}\]

    So, the derivative of y = x\sqrt{x+1} with respect to x is:

        \[\frac{d}{dx}[x\sqrt{x+1}] &= \frac{x}{2\sqrt{x+1}} + \sqrt{x+1}\]

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