Find the derivative of the function \( T(x)=\tan \left(\frac{2 x+7}{3 x-1}\right) \)

To find the derivative of the function \( T(x) = \tan \left(\frac{2x + 7}{3x - 1}\right) \), we can use the chain rule and the quotient rule. 1. **Chain Rule**: The derivative of \( \tan(u) \) is \( \sec^2(u) \cdot u' \), where \( u = \frac{2x + 7}{3x - 1} \). 2. **Quotient Rule**: To find \( u' \), apply the quotient rule: if \( u = \frac{f(x)}{g(x)} \), then \( u' = \frac{f'g - fg'}{g^2} \). Here, \( f(x) = 2x + 7 \) and \( g(x) = 3x - 1 \). - \( f'(x) = 2 \) - \( g'(x) = 3 \) Now applying the quotient rule: \[ u' = \frac{(2)(3x - 1) - (2x + 7)(3)}{(3x - 1)^2} \] Simplifying this gives: \[ u' = \frac{6x - 2 - 6x - 21}{(3x - 1)^2} = \frac{-23}{(3x - 1)^2} \] 3. Now substituting \( u' \) back into the derivative of \( T(x) \): \[ T'(x) = \sec^2\left(\frac{2x + 7}{3x - 1}\right) \cdot \left( \frac{-23}{(3x - 1)^2} \right) \] Finally, you can express the derivative as: \[ T'(x) = -23 \cdot \frac{\sec^2\left(\frac{2x + 7}{3x - 1}\right)}{(3x - 1)^2} \] So the derivative of \( T(x) = \tan \left(\frac{2 x+7}{3 x-1}\right) \) is: \[ T'(x) = -23 \cdot \frac{\sec^2\left(\frac{2x + 7}{3x - 1}\right)}{(3x - 1)^2} \]