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Miscellaneous Exercise · Q7

Q.Find the derivative of 1ax2+bx+c\dfrac{1}{ax^2 + bx + c}.

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The derivative of 1ax2+bx+c\frac{1}{ax^2 + bx + c} is found using the chain rule (or quotient rule) and is −2ax+b(ax2+bx+c)2-\frac{2ax + b}{(ax^2 + bx + c)^2}.

The function we have is a reciprocal of a quadratic. When you see something like 1something\frac{1}{\text{something}}, the instinct should be to think of it as (something)−1(\text{something})^{-1}. This is a classic composition of functions: the outer function is u−1u^{-1} and the inner function is u=ax2+bx+cu = ax^2 + bx + c.

The chain rule says: derivative of f(g(x))f(g(x)) is f′(g(x))⋅g′(x)f'(g(x)) \cdot g'(x). Here, f(u)=u−1f(u) = u^{-1}, so f′(u)=−u−2f'(u) = -u^{-2}. Multiply by the derivative of the inner function g′(x)=2ax+bg'(x) = 2ax + b, and you get the answer directly.

You could also use the quotient rule directly on 1ax2+bx+c\frac{1}{ax^2 + bx + c}, treating the numerator as 11 and denominator as D(x)D(x). The quotient rule says ddx(ND)=N′D−ND′D2\frac{d}{dx}\left(\frac{N}{D}\right) = \frac{N'D - ND'}{D^2}. With N=1N = 1 and N′=0N' = 0, this simplifies to −D′D2-\frac{D'}{D^2}, which is exactly the same as the chain rule result. Both paths lead to the same place — pick whichever feels more natural.

Let's work through it step by step.

  1. Identify the structure.

    Write y=1ax2+bx+cy = \frac{1}{ax^2 + bx + c}. This is y=(ax2+bx+c)−1y = (ax^2 + bx + c)^{-1}.

  2. Apply the chain rule.

    Let u=ax2+bx+cu = ax^2 + bx + c. Then y=u−1y = u^{-1}.

    dydu=−u−2\frac{dy}{du} = -u^{-2} and dudx=2ax+b\frac{du}{dx} = 2ax + b.

  3. Multiply the derivatives.

    dydx=dydu⋅dudx=(−u−2)⋅(2ax+b)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (-u^{-2}) \cdot (2ax + b). …

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