Differentiate with respect to $x$: $\dfrac{1}{ax^2 + bx + c}$.

> [!TLDR] > To differentiate $\dfrac{1}{ax^2 + bx + c}$, we apply the Quotient Rule, treating the numerator as $u=1$ and the denominator as $v=ax^2+bx+c$, which yields the derivative $\boxed{-\dfrac{2ax + b}{(ax^2 + bx + c)^2}}$. We need to find the derivative of the function $f(x) = \dfrac{1}{ax^2 + bx + c}$ with respect to $x$. This function is presented as a quotient of two expressions, which immediately suggests using the Quotient Rule for differentiation. The Quotient Rule is a fundamental tool for finding the derivative of a function that is expressed as a ratio of two other differentiable functions. If we have a function $f(x) = \dfrac{u(x)}{v(x)}$, where $u(x)$ is the numerator and $v(x)$ is the denominator, then its derivative $f'(x)$ is given by: > [!FORMULA] > $$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} $$ > where $u'$ is the derivative of $u$ with respect to $x$, and $v'$ is the derivative of $v$ with respect to $x$. Let's apply this rule step-by-step to our function. 1. **Identify the numerator and denominator functions.** For $f(x) = \dfrac{1}{ax^2 + bx + c}$: Let $u(x) = 1$ (the numerator). Let $v(x) = ax^2 + bx + c$ (the denominator). 2. **Differentiate the numerator and denominator functions.** We need to find $u'(x)$ and $v'(x)$. * The derivative of a constant is always zero. So, $u'(x) = \dfrac{d}{dx}(1) = 0$. * For $v(x) = ax^2 + bx + c$, we differentiate term by term using the power rule and the constant multiple rule: $\dfrac{d}{dx}(ax^2) = a \cdot 2x = 2ax$ $\dfrac{d}{dx}(bx) = b \cdot 1 = b$ $\dfrac{d}{dx}(c) = 0$ So, $v'(x) = \dfrac{d}{dx}(ax^2 + bx + c) = 2ax + b$. 3. **Substitute these into the Quotient Rule formula.** Now we have all the components: $u = 1$ $u' = 0$ $v = ax^2 + bx + c$ $v' = 2ax + b$ Plugging these into the formula $\dfrac{u'v - uv'}{v^2}$: $$ \frac{d}{dx}\left(\frac{1}{ax^2 + bx + c}\right) = \frac{(0)(a

Differentiate with respect to x: (1)/(ax² + bx + c).

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Concept understanding — Rules of Differentiation

Computing every derivative directly from the limit definition ("from first principle") is correct but impractical for anything beyond the simplest functions. The five rules below, proved once from the definition, make differentiation a mechanical, algebraic process for any combination of known derivatives — no further limit computation is needed once they are established. Throughout, $u=u(x)$ , $v=v(x)$ are differentiable functions of $x$ , and $k$ is a constant.

Sum Rule (Theorem 10.2). $\dfrac{d}{dx}(u+v)=\dfrac{du}{dx}+\dfrac{dv}{dx}$ — differentiate term by term. Extends to any finite sum.

Product Rule (Theorem 10.3). $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ , i.e. $(uv)'=uv'+vu'$ — "first times derivative of second, plus second times derivative of first." Extends to three or more factors, e.g. $(uvw)'=u'vw+uv'w+uvw'$ : differentiate one factor at a time, summing over which factor is differentiated.

Quotient Rule (Theorem 10.4). For $v\ne0$ , $\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{v\,u'-u\,v'}{v^2}$ — "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared."

Chain Rule (Theorem 10.5). If $y=f(u)$ and $u=g(x)$ are both differentiable, so $y=f(g(x))$ , then

\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = f'(g(x))\,g'(x)

— differentiate the outer function $f$ with respect to its argument $u=g(x)$ (the inner function), then multiply by the derivative of the inner function. This is the single most-used rule in the chapter: every composite function (a power of an expression, $\sin$ of an expression, $e^{\text{expression}}$ , and so on) needs it.

Constant Multiple Rule (Theorem 10.6). $\dfrac{d}{dx}\big[k\,f(x)\big] = k\,f'(x)$ — a constant factor simply carries through differentiation unchanged. …

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To differentiate $\dfrac{1}{ax^2 + bx + c}$ , we apply the Quotient Rule, treating the numerator as $u=1$ and the denominator as $v=ax^2+bx+c$ , which yields the derivative $\boxed{-\dfrac{2ax + b}{(ax^2 + bx + c)^2}}$ .

We need to find the derivative of the function $f(x) = \dfrac{1}{ax^2 + bx + c}$ with respect to $x$ . This function is presented as a quotient of two expressions, which immediately suggests using the Quotient Rule for differentiation.

The Quotient Rule is a fundamental tool for finding the derivative of a function that is expressed as a ratio of two other differentiable functions. If we have a function $f(x) = \dfrac{u(x)}{v(x)}$ , where $u(x)$ is the numerator and $v(x)$ is the denominator, then its derivative $f'(x)$ is given by:

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

Let's apply this rule step-by-step to our function.

Identify the numerator and denominator functions.

For $f(x) = \dfrac{1}{ax^2 + bx + c}$ :

Let $u(x) = 1$ (the numerator).

Let $v(x) = ax^2 + bx + c$ (the denominator).
Differentiate the numerator and denominator functions.

We need to find $u'(x)$ and $v'(x)$ .
- The derivative of a constant is always zero. So, $u'(x) = \dfrac{d}{dx}(1) = 0$ .
- For $v(x) = ax^2 + bx + c$ , we differentiate term by term using the power rule and the constant multiple rule: $\dfrac{d}{dx}(ax^2) = a \cdot 2x = 2ax$ $\dfrac{d}{dx}(bx) = b \cdot 1 = b$ $\dfrac{d}{dx}(c) = 0$ So, $v'(x) = \dfrac{d}{dx}(ax^2 + bx + c) = 2ax + b$ .
Substitute these into the Quotient Rule formula.

Now we have all the components:

$u = 1$

$u' = 0$

$v = ax^2 + bx + c$

$v' = 2ax + b$

Plugging these into the formula $\dfrac{u'v - uv'}{v^2}$ :

\frac{d}{dx}\left(\frac{1}{ax^2 + bx + c}\right) = \frac{(0)(ax^2 + bx + c) - (1)(2ax + b)}{(ax^2 + bx + c)^2}

Simplify the expression. The first term in the numerator, $(0)(ax^2 + bx + c)$ , becomes $0$ . …

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Q.Differentiate with respect to xxx: 1ax2+bx+c\dfrac{1}{ax^2 + bx + c}ax2+bx+c1​.

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Q.Differentiate with respect to $x$ : $\dfrac{1}{ax^2 + bx + c}$ .

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