Questions: Find the derivative of the function. y=3x/(√(x^2+2)) y'(x)=□

Transcript text: Find the derivative of the function. \[ \begin{array}{c} y=\frac{3 x}{\sqrt{x^{2}+2}} \\ y^{\prime}(x)=\square \end{array} \]

Solution

Solution Steps

To find the derivative of the function \( y = \frac{3x}{\sqrt{x^2 + 2}} \), we can use the quotient rule. The quotient rule states that if you have a function in the form \( \frac{u(x)}{v(x)} \), its derivative is given by \( \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). Here, \( u(x) = 3x \) and \( v(x) = \sqrt{x^2 + 2} \).

Step 1: Define the Function

We start with the function: \[ y = \frac{3x}{\sqrt{x^2 + 2}} \]

Step 2: Apply the Quotient Rule

To find the derivative \( y'(x) \), we use the quotient rule: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] where \( u = 3x \) and \( v = \sqrt{x^2 + 2} \).

Step 3: Compute \( u' \) and \( v' \)

First, compute the derivatives of \( u \) and \( v \): \[ u' = \frac{d}{dx}(3x) = 3 \] \[ v = (x^2 + 2)^{1/2} \implies v' = \frac{d}{dx} (x^2 + 2)^{1/2} = \frac{1}{2}(x^2 + 2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 2}} \]

Step 4: Substitute into the Quotient Rule

Substitute \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula: \[ y' = \frac{3 \cdot \sqrt{x^2 + 2} - 3x \cdot \frac{x}{\sqrt{x^2 + 2}}}{(x^2 + 2)} \]

Step 5: Simplify the Expression

Simplify the numerator: \[ y' = \frac{3\sqrt{x^2 + 2} - \frac{3x^2}{\sqrt{x^2 + 2}}}{x^2 + 2} = \frac{3(x^2 + 2) - 3x^2}{(x^2 + 2)^{3/2}} = \frac{6}{(x^2 + 2)^{3/2}} \]