Problem 25 - Single Variable Calculus

Question

Find the derivative of each function. \(f(x)=\frac{x+\sqrt{3 x}}{3 x-1}\)

Step-by-Step Solution

Answer

The derivative of the given function \(f(x)=\frac{x+\sqrt{3 x}}{3 x-1}\) is \(f'(x) = \frac{-1}{(3x-1)^2}\).

Step 1: Identify g(x) and h(x)

We have a function in the form of a fraction: \(f(x)=\frac{x+\sqrt{3 x}}{3 x-1}\). We can identify \(g(x)\) as the function in the numerator and \(h(x)\) as the function in the denominator: - \(g(x) = x+\sqrt{3x}\) - \(h(x) = 3x-1\)

Step 2: Find g'(x) and h'(x)

Next, we need to find the derivatives of these two functions: - To find the derivative of \(g(x) = x+\sqrt{3x}\), notice that we need to apply the chain rule for the second term: \(g'(x) = \frac{d}{dx}(x+\sqrt{3x}) = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{3x}) = 1 + \frac{1}{2}(3x)^{-\frac{1}{2}}\cdot(3) = 1 + \frac{3}{2\sqrt{3x}}\) - To find the derivative of \(h(x) = 3x-1\): \(h'(x) = \frac{d}{dx}(3x-1) = 3\)

Step 3: Apply the Quotient Rule

Now that we have found the derivatives of \(g(x)\) and \(h(x)\), we can apply the quotient rule to find the derivative of \(f(x)\): $$ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} $$ Plugging in the values we found in the previous steps: $$ f'(x) = \frac{(1 + \frac{3}{2\sqrt{3x}})(3x-1) - (x+\sqrt{3x})(3)}{(3x-1)^2} $$

Step 4: Simplify the result

Finally, we'll simplify the result as much as possible $$ f'(x) = \frac{(3x + \frac{9}{2\sqrt{3x}} - 1 - \frac{3}{2\sqrt{3x}}) - (3x+3\sqrt{3x})}{(3x-1)^2} $$ $$ f'(x) = \frac{-1 + 3\sqrt{3x} - 3\sqrt{3x}}{(3x-1)^2} $$ Thus, the derivative of the given function is: $$ f'(x) = \frac{-1}{(3x-1)^2} $$