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Find the derivative of y = (x2 + 7x) / (x+2) using the quotient rule.

Mathematics

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Answer

dydx=x2+4x+14(x+2)2\frac{dy}{dx} = \frac{x^2 + 4x + 14}{(x+2)^2}

Graph of y = (x² + 7x)/(x + 2)

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Key points of y = (x² + 7x)/(x + 2)
Pointxy
x-intercept−70
x-intercept00

Step-by-step solution

5 steps

To find dydx\frac{dy}{dx} for the given function, we will use the quotient rule.

The function is y=x2+7xx+2y = \frac{x^2 + 7x}{x+2}.
Let u=x2+7xu = x^2 + 7x and v=x+2v = x+2.

  1. Find the derivatives of uu and vv with respect to xx.

    dudx=ddx(x2+7x)=2x+7\frac{du}{dx} = \frac{d}{dx}(x^2 + 7x) = 2x + 7 dvdx=ddx(x+2)=1\frac{dv}{dx} = \frac{d}{dx}(x+2) = 1
  2. Apply the quotient rule, which states that if y=uvy = \frac{u}{v}, then dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}.

    dydx=(x+2)(2x+7)−(x2+7x)(1)(x+2)2\frac{dy}{dx} = \frac{(x+2)(2x+7) - (x^2+7x)(1)}{(x+2)^2}
  3. Expand the numerator.

    (x+2)(2x+7)−(x2+7x)(1)=(2x2+7x+4x+14)−(x2+7x)(x+2)(2x+7) - (x^2+7x)(1) = (2x^2 + 7x + 4x + 14) - (x^2 + 7x) =2x2+11x+14−x2−7x= 2x^2 + 11x + 14 - x^2 - 7x
  4. Combine like terms in the numerator.

    (2x2−x2)+(11x−7x)+14=x2+4x+14(2x^2 - x^2) + (11x - 7x) + 14 = x^2 + 4x + 14
  5. Write the final expression for dydx\frac{dy}{dx}.

    dydx=x2+4x+14(x+2)2\frac{dy}{dx} = \frac{x^2 + 4x + 14}{(x+2)^2}
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