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Differentiate y = sqrt(x)(x2+2)3 using the product rule.

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Answer

dydx=(x2+2)2(13x2+2)2x\frac{dy}{dx} = \frac{(x^2+2)^2 (13x^2+2)}{2\sqrt{x}}

Graph of y = √x(x² + 2)³

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Step-by-step solution

5 steps

To differentiate y=x(x2+2)3y = \sqrt{x}(x^2+2)^3 using the product rule, we use the formula dydx=u′v+uv′\frac{dy}{dx} = u'v + uv'.

  1. Identify uu and vv.

    Let u=x=x1/2u = \sqrt{x} = x^{1/2}.
    Let v=(x2+2)3v = (x^2+2)^3.

  2. Find the derivative of uu, denoted as u′u'.

    u′=ddx(x1/2)=12x1/2−1=12x−1/2=12xu' = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}

  3. Find the derivative of vv, denoted as v′v'.

    This requires the chain rule.
    Let w=x2+2w = x^2+2. Then v=w3v = w^3.
    Using the chain rule, dvdx=dvdw⋅dwdx\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx}.

    dvdw=ddw(w3)=3w2\frac{dv}{dw} = \frac{d}{dw}(w^3) = 3w^2 dwdx=ddx(x2+2)=2x\frac{dw}{dx} = \frac{d}{dx}(x^2+2) = 2x

    So,

    v′=3(x2+2)2⋅(2x)=6x(x2+2)2v' = 3(x^2+2)^2 \cdot (2x) = 6x(x^2+2)^2

  4. Apply the product rule formula dydx=u′v+uv′\frac{dy}{dx} = u'v + uv'.

    dydx=(12x−1/2)(x2+2)3+(x1/2)(6x(x2+2)2)\frac{dy}{dx} = \left(\frac{1}{2}x^{-1/2}\right)(x^2+2)^3 + (x^{1/2})\left(6x(x^2+2)^2\right)

  5. Simplify the expression.

    Rewrite x−1/2x^{-1/2} as 1x\frac{1}{\sqrt{x}} and x1/2x^{1/2} as x\sqrt{x}.

    dydx=(x2+2)32x+6xx(x2+2)2\frac{dy}{dx} = \frac{(x^2+2)^3}{2\sqrt{x}} + 6x\sqrt{x}(x^2+2)^2

    Factor out common terms, which are (x2+2)2(x^2+2)^2 and 12x\frac{1}{2\sqrt{x}}.

    dydx=(x2+2)22x[(x2+2)+2x⋅6xx]\frac{dy}{dx} = \frac{(x^2+2)^2}{2\sqrt{x}} \left[ (x^2+2) + 2\sqrt{x} \cdot 6x\sqrt{x} \right]

    Simplify the term 2x⋅6xx2\sqrt{x} \cdot 6x\sqrt{x}:

    2x⋅6xx=12x(x)2=12x⋅x=12x22\sqrt{x} \cdot 6x\sqrt{x} = 12x (\sqrt{x})^2 = 12x \cdot x = 12x^2

    Substitute this back into the expression:

    dydx=(x2+2)22x[(x2+2)+12x2]\frac{dy}{dx} = \frac{(x^2+2)^2}{2\sqrt{x}} \left[ (x^2+2) + 12x^2 \right]

    Combine terms inside the brackets:

    dydx=(x2+2)22x[x2+2+12x2]\frac{dy}{dx} = \frac{(x^2+2)^2}{2\sqrt{x}} \left[ x^2+2+12x^2 \right] dydx=(x2+2)2(13x2+2)2x\frac{dy}{dx} = \frac{(x^2+2)^2 (13x^2+2)}{2\sqrt{x}}
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