(iv) Let `y= (1)/( ax^(2) + bx + c)` then find dy/dx

Answer

To find the derivative of the function \( y = \frac{1}{ax^2 + bx + c} \), we will use the chain rule and the power rule of differentiation. Here’s the step-by-step solution:

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

### Step 2: Differentiate using the chain rule
Now, we will differentiate \( y \) with respect to \( x \). Using the chain rule, we have:
\[ \frac{dy}{dx} = -1 \cdot (ax^2 + bx + c)^{-2} \cdot \frac{d}{dx}(ax^2 + bx + c) \]

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

### Step 4: Substitute back into the derivative
Now, substituting this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -1 \cdot (ax^2 + bx + c)^{-2} \cdot (2ax + b) \]

### Step 5: Simplify the expression
Thus, we can write:
\[ \frac{dy}{dx} = -\frac{2ax + b}{(ax^2 + bx + c)^2} \]

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

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