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Find the derivatives of the following functions: i) y = (2+3x2−x)/(sqrt(1+x)), ii) y = (x2−2)/(x3-2)

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3 parts
i

9x2+11x−42(1+x)3/2\frac{9x^2+11x-4}{2(1+x)^{3/2}}

ii

−x4+6x2−4x(x3−2)2\frac{-x^4+6x^2-4x}{(x^3-2)^2}

iii

−8x(2x2−1)2\frac{-8x}{(2x^2-1)^2}

Step-by-step solution

9 steps

Here are the derivatives for the given functions:

Part i· 3 steps

To find the derivative of y=2+3x2−x1+xy = \frac{2+3x^2-x}{\sqrt{1+x}} with respect to xx, we use the quotient rule ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.

  1. Identify uu and vv and their derivatives.

    Let u=2+3x2−xu = 2+3x^2-x, so u′=6x−1u' = 6x-1.
    Let v=1+x=(1+x)1/2v = \sqrt{1+x} = (1+x)^{1/2}, so v′=12(1+x)−1/2=121+xv' = \frac{1}{2}(1+x)^{-1/2} = \frac{1}{2\sqrt{1+x}}.

  2. Apply the quotient rule.

    dydx=(6x−1)1+x−(2+3x2−x)(121+x)(1+x)2\frac{dy}{dx} = \frac{(6x-1)\sqrt{1+x} - (2+3x^2-x)\left(\frac{1}{2\sqrt{1+x}}\right)}{(\sqrt{1+x})^2}

  3. Simplify the expression.

    dydx=2(6x−1)(1+x)−(2+3x2−x)21+x1+x\frac{dy}{dx} = \frac{\frac{2(6x-1)(1+x) - (2+3x^2-x)}{2\sqrt{1+x}}}{1+x} dydx=2(6x2+5x−1)−(2+3x2−x)21+x(1+x)\frac{dy}{dx} = \frac{2(6x^2+5x-1) - (2+3x^2-x)}{2\sqrt{1+x}(1+x)} dydx=12x2+10x−2−2−3x2+x2(1+x)3/2\frac{dy}{dx} = \frac{12x^2+10x-2 - 2-3x^2+x}{2(1+x)^{3/2}} dydx=9x2+11x−42(1+x)3/2\frac{dy}{dx} = \boxed{\frac{9x^2+11x-4}{2(1+x)^{3/2}}}

Part ii

−x4+6x2−4x(x3−2)2\frac{-x^4+6x^2-4x}{(x^3-2)^2}

To find the derivative of y=x2−2x3−2y = \frac{x^2-2}{x^3-2} with respect to xx, we use the quotient rule.

  1. Identify uu and vv and their derivatives.

    Let u=x2−2u = x^2-2, so u′=2xu' = 2x.
    Let v=x3−2v = x^3-2, so v′=3x2v' = 3x^2.

  2. Apply the quotient rule.

    dydx=(2x)(x3−2)−(x2−2)(3x2)(x3−2)2\frac{dy}{dx} = \frac{(2x)(x^3-2) - (x^2-2)(3x^2)}{(x^3-2)^2}

  3. Simplify the expression.

    dydx=2x4−4x−(3x4−6x2)(x3−2)2\frac{dy}{dx} = \frac{2x^4-4x - (3x^4-6x^2)}{(x^3-2)^2} dydx=2x4−4x−3x4+6x2(x3−2)2\frac{dy}{dx} = \frac{2x^4-4x - 3x^4+6x^2}{(x^3-2)^2} dydx=−x4+6x2−4x(x3−2)2\frac{dy}{dx} = \boxed{\frac{-x^4+6x^2-4x}{(x^3-2)^2}}

Part iii

−8x(2x2−1)2\frac{-8x}{(2x^2-1)^2}

To find the derivative of y=8x2+48x2−4y = \frac{8x^2+4}{8x^2-4} with respect to xx, we use the quotient rule.

  1. Identify uu and vv and their derivatives.

    Let u=8x2+4u = 8x^2+4, so u′=16xu' = 16x.
    Let v=8x2−4v = 8x^2-4, so v′=16xv' = 16x.

  2. Apply the quotient rule.

    dydx=(16x)(8x2−4)−(8x2+4)(16x)(8x2−4)2\frac{dy}{dx} = \frac{(16x)(8x^2-4) - (8x^2+4)(16x)}{(8x^2-4)^2}

  3. Simplify the expression.

    dydx=16x[(8x2−4)−(8x2+4)](8x2−4)2\frac{dy}{dx} = \frac{16x[(8x^2-4) - (8x^2+4)]}{(8x^2-4)^2} dydx=16x[8x2−4−8x2−4](8x2−4)2\frac{dy}{dx} = \frac{16x[8x^2-4 - 8x^2-4]}{(8x^2-4)^2} dydx=16x(−8)(8x2−4)2\frac{dy}{dx} = \frac{16x(-8)}{(8x^2-4)^2} dydx=−128x[4(2x2−1)]2\frac{dy}{dx} = \frac{-128x}{[4(2x^2-1)]^2} dydx=−128x16(2x2−1)2\frac{dy}{dx} = \frac{-128x}{16(2x^2-1)^2} dydx=−8x(2x2−1)2\frac{dy}{dx} = \boxed{\frac{-8x}{(2x^2-1)^2}}
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