Find the derivative of the function \( T(x)=\tan \left(\frac{4 x+7}{5 x-4}\right) \)

To find the derivative of the function \( T(x) = \tan\left(\frac{4x+7}{5x-4}\right) \), we will apply the chain rule and the quotient rule. 1. Let \( u = \frac{4x+7}{5x-4} \), then \( T(x) = \tan(u) \). 2. The derivative of \( \tan(u) \) is \( \sec^2(u) \cdot \frac{du}{dx} \). 3. Next, we need to find \( \frac{du}{dx} \) using the quotient rule: if \( u = \frac{f(x)}{g(x)} \), then \( \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \), where \( f(x) = 4x + 7 \) and \( g(x) = 5x - 4 \). Now we compute: - \( f'(x) = 4 \) - \( g'(x) = 5 \) Using the quotient rule: \[ \frac{du}{dx} = \frac{(4)(5x - 4) - (4x + 7)(5)}{(5x - 4)^2} \] Simplifying the numerator: \[ = \frac{20x - 16 - (20x + 35)}{(5x - 4)^2} \] \[ = \frac{20x - 16 - 20x - 35}{(5x - 4)^2} = \frac{-51}{(5x - 4)^2} \] Now, substituting \( u \) and \( \frac{du}{dx} \) back into the derivative of \( T(x) \): \[ T'(x) = \sec^2\left(\frac{4x + 7}{5x - 4}\right) \cdot \frac{-51}{(5x - 4)^2} \] Thus, the derivative of the function \( T(x) \) is: \[ T'(x) = -\frac{51 \sec^2\left(\frac{4x + 7}{5x - 4}\right)}{(5x - 4)^2} \]