If f(1÷x) = (3x + 1) ÷ (2x - 1) then dy÷dx when x = 3 equals what

To solve for \(\frac{dy}{dx}\) when \(x = 3\) given that \(f\left(\frac{1}{x}\right) = \frac{3x + 1}{2x - 1}\), we need to follow these steps:

### Step 1: Express \(y\) in terms of \(x\)
Let \(y = f\left(\frac{1}{x}\right)\).

Given \(f\left(\frac{1}{x}\right) = \frac{3x + 1}{2x - 1}\), we have:
\[
y = \frac{3x + 1}{2x - 1}
\]

### Step 2: Differentiate \(y\) with respect to \(x\)
We'll differentiate \(y\) with respect to \(x\) using the quotient rule.

The quotient rule states that if \(y = \frac{u(x)}{v(x)}\), then:
\[
\frac{dy}{dx} = \frac{v(x) \cdot \frac{du}{dx} - u(x) \cdot \frac{dv}{dx}}{[v(x)]^2}
\]

Here, \(u(x) = 3x + 1\) and \(v(x) = 2x - 1\).

First, find \(\frac{du}{dx}\) and \(\frac{dv}{dx}\):
\[
\frac{du}{dx} = 3
\]
\[
\frac{dv}{dx} = 2
\]

Now apply the quotient rule:
\[
\frac{dy}{dx} = \frac{(2x - 1) \cdot 3 - (3x + 1) \cdot 2}{(2x - 1)^2}
\]

Simplify the numerator:
\[
\frac{dy}{dx} = \frac{6x - 3 - (6x + 2)}{(2x - 1)^2}
\]
\[
\frac{dy}{dx} = \frac{6x - 3 - 6x - 2}{(2x - 1)^2}
\]
\[
\frac{dy}{dx} = \frac{-5}{(2x - 1)^2}
\]

### Step 3: Evaluate \(\frac{dy}{dx}\) at \(x = 3\)
Substitute \(x = 3\) into the expression for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{-5}{(2(3) - 1)^2} = \frac{-5}{(6 - 1)^2} = \frac{-5}{25} = -\frac{1}{5}
\]

So, \(\frac{dy}{dx}\) when \(x = 3\) equals \(-\frac{1}{5}\).

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you apply the chain rule in differentiation problems?
2. What are the conditions under which the quotient rule is used?
3. How can we differentiate implicit functions?
4. Can you find the second derivative for this function?
5. What happens to the derivative when the denominator equals zero?

**Tip:** Always simplify the expression before substituting values to avoid unnecessary complexity in calculations.

Discover how to find the derivative dy/dx at x = 3 for the function f(1/x) = (3x + 1) / (2x - 1) using the quotient rule and differentiation techniques. This problem involves advanced differentiation concepts suitable for high school students.

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