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Example Find dy/dx for questions a) to m)

Mathematics

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Answer

dydx=−x−52x+1(x−3)2\frac{dy}{dx} = \frac{-x-5}{2\sqrt{x+1}(x-3)^2}

Step-by-step solution

5 steps

Ili kutatua tatizo (m), tutatumia kanuni ya quotient na kanuni ya mnyororo.

Tatizo ni:

y=x+1x−3y = \frac{\sqrt{x+1}}{x-3}
  1. Tambua uu na vv.

    Hapa, u=x+1=(x+1)1/2u = \sqrt{x+1} = (x+1)^{1/2} na v=x−3v = x-3.

  2. Tafuta dudx\frac{du}{dx} na dvdx\frac{dv}{dx}.

    Kwa u=(x+1)1/2u = (x+1)^{1/2}:

    dudx=12(x+1)12−1⋅ddx(x+1)\frac{du}{dx} = \frac{1}{2}(x+1)^{\frac{1}{2}-1} \cdot \frac{d}{dx}(x+1) dudx=12(x+1)−1/2⋅1\frac{du}{dx} = \frac{1}{2}(x+1)^{-1/2} \cdot 1 dudx=12x+1\frac{du}{dx} = \frac{1}{2\sqrt{x+1}}

    Kwa v=x−3v = x-3:

    dvdx=1\frac{dv}{dx} = 1
  3. Tumia kanuni ya quotient.

    Kanuni ya quotient inasema: dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}.
    Badilisha maadili ya u,v,dudx,dvdxu, v, \frac{du}{dx}, \frac{dv}{dx}:

    dydx=(x−3)(12x+1)−(x+1)(1)(x−3)2\frac{dy}{dx} = \frac{(x-3)\left(\frac{1}{2\sqrt{x+1}}\right) - (\sqrt{x+1})(1)}{(x-3)^2}
  4. Rahisisha nambari (numerator).

    Nambari ni: x−32x+1−x+1\frac{x-3}{2\sqrt{x+1}} - \sqrt{x+1}
    Ili kuchanganya maneno haya, tafuta denominator ya kawaida, ambayo ni 2x+12\sqrt{x+1}:

    Nambari=x−32x+1−x+1⋅2x+12x+1\text{Nambari} = \frac{x-3}{2\sqrt{x+1}} - \frac{\sqrt{x+1} \cdot 2\sqrt{x+1}}{2\sqrt{x+1}} Nambari=x−3−2(x+1)2x+1\text{Nambari} = \frac{x-3 - 2(x+1)}{2\sqrt{x+1}} Nambari=x−3−2x−22x+1\text{Nambari} = \frac{x-3 - 2x - 2}{2\sqrt{x+1}} Nambari=−x−52x+1\text{Nambari} = \frac{-x-5}{2\sqrt{x+1}}
  5. Badilisha nambari iliyorahisishwa kwenye usemi wa dydx\frac{dy}{dx}.

    dydx=−x−52x+1(x−3)2\frac{dy}{dx} = \frac{\frac{-x-5}{2\sqrt{x+1}}}{(x-3)^2} dydx=−x−52x+1(x−3)2\frac{dy}{dx} = \frac{-x-5}{2\sqrt{x+1}(x-3)^2}
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