DIfferentiate the following w.r.t x :`(1)/(ax^2+bx+c)`

Answer

To differentiate the function \( y = \frac{1}{ax^2 + bx + c} \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the function
We start by rewriting the function in a form that is easier to differentiate:
\[ y = (ax^2 + bx + c)^{-1} \]

### Step 2: Apply the chain rule
Using the chain rule, we differentiate \( y \):
\[ \frac{dy}{dx} = -1 \cdot (ax^2 + bx + c)^{-2} \cdot \frac{d}{dx}(ax^2 + bx + c) \]

### Step 3: Differentiate the inner function
Now, we need to differentiate the inner function \( ax^2 + bx + c \):
\[ \frac{d}{dx}(ax^2 + bx + c) = 2ax + b \]

### Step 4: Substitute back into the derivative
Substituting the derivative of the inner function back into our expression, we have:
\[ \frac{dy}{dx} = -1 \cdot (ax^2 + bx + c)^{-2} \cdot (2ax + b) \]

### Step 5: Simplify the expression
This gives us:
\[ \frac{dy}{dx} = -\frac{2ax + b}{(ax^2 + bx + c)^2} \]

### Final Result
Thus, the derivative of \( y = \frac{1}{ax^2 + bx + c} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{2ax + b}{(ax^2 + bx + c)^2} \]

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