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Find the derivative of y = (2x2 + 1) / (x5 + 2x) using the quotient rule.

Mathematics

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Answer

dydx=−6x6−5x4+4x2−2(x5+2x)2\frac{dy}{dx} = \frac{-6x^6 - 5x^4 + 4x^2 - 2}{(x^5 + 2x)^2}

Graph of y = (2x² + 1)/(x⁵ + 2x)

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Key points of y = (2x² + 1)/(x⁵ + 2x)
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Step-by-step solution

4 steps

To find the derivative of y=2x2+1x5+2xy = \frac{2x^2 + 1}{x^5 + 2x}, we will use the quotient rule, which states that if y=uvy = \frac{u}{v}, then dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.

  1. Identify uu and vv and find their derivatives.

    Let u=2x2+1u = 2x^2 + 1.
    Then u′=ddx(2x2+1)=4xu' = \frac{d}{dx}(2x^2 + 1) = 4x.
    Let v=x5+2xv = x^5 + 2x.
    Then v′=ddx(x5+2x)=5x4+2v' = \frac{d}{dx}(x^5 + 2x) = 5x^4 + 2.

  2. Apply the quotient rule formula.

    dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

    Substitute the expressions for u,v,u′u, v, u', and v′v':

    dydx=(4x)(x5+2x)−(2x2+1)(5x4+2)(x5+2x)2\frac{dy}{dx} = \frac{(4x)(x^5 + 2x) - (2x^2 + 1)(5x^4 + 2)}{(x^5 + 2x)^2}

  3. Expand the terms in the numerator.

    First term: (4x)(x5+2x)=4x⋅x5+4x⋅2x=4x6+8x2(4x)(x^5 + 2x) = 4x \cdot x^5 + 4x \cdot 2x = 4x^6 + 8x^2.
    Second term: (2x2+1)(5x4+2)=2x2(5x4)+2x2(2)+1(5x4)+1(2)=10x6+4x2+5x4+2(2x^2 + 1)(5x^4 + 2) = 2x^2(5x^4) + 2x^2(2) + 1(5x^4) + 1(2) = 10x^6 + 4x^2 + 5x^4 + 2.

  4. Substitute the expanded terms back into the numerator and simplify.

    dydx=(4x6+8x2)−(10x6+5x4+4x2+2)(x5+2x)2\frac{dy}{dx} = \frac{(4x^6 + 8x^2) - (10x^6 + 5x^4 + 4x^2 + 2)}{(x^5 + 2x)^2}

    Distribute the negative sign in the numerator:

    dydx=4x6+8x2−10x6−5x4−4x2−2(x5+2x)2\frac{dy}{dx} = \frac{4x^6 + 8x^2 - 10x^6 - 5x^4 - 4x^2 - 2}{(x^5 + 2x)^2}

    Combine like terms in the numerator:

    dydx=(4x6−10x6)−5x4+(8x2−4x2)−2(x5+2x)2\frac{dy}{dx} = \frac{(4x^6 - 10x^6) - 5x^4 + (8x^2 - 4x^2) - 2}{(x^5 + 2x)^2} dydx=−6x6−5x4+4x2−2(x5+2x)2\frac{dy}{dx} = \frac{-6x^6 - 5x^4 + 4x^2 - 2}{(x^5 + 2x)^2}
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