Differentiate with respect to \( x \) if \( y=2 \log x-\tan ^{2} x-3 r^{2 x}+x^{-2}-3^{-2 x}+9 x+6 \)

To differentiate the function \( y = 2 \log x - \tan^2 x - 3 r^{2x} + x^{-2} - 3^{-2x} + 9x + 6 \) with respect to \( x \), we need to find the derivative of each term individually. 1. The derivative of \( 2 \log x \) is \( \frac{2}{x} \). 2. The derivative of \( \tan^2 x \) can be found using the chain rule: \( 2 \tan x \sec^2 x \). 3. The derivative of \( -3 r^{2x} \) is \( -3 \cdot 2 \ln(r) \cdot r^{2x} \) since we differentiate with respect to \( x \). 4. The derivative of \( x^{-2} \) is \( -2x^{-3} \). 5. The derivative of \( -3^{-2x} \) can be found similarly: \( -2 \ln(3) \cdot 3^{-2x} \). 6. The derivative of \( 9x \) is \( 9 \). 7. The derivative of the constant \( 6 \) is \( 0 \). Now, putting it all together: \[ \frac{dy}{dx} = \frac{2}{x} - 2 \tan x \sec^2 x - 6 \ln(r) \cdot 3^{2x} + 2x^{-3} + 2 \ln(3) \cdot 3^{-2x} + 9 \] So, the final derivative is: \[ \frac{dy}{dx} = \frac{2}{x} - 2 \tan x \sec^2 x - 6 \ln(r) \cdot 3^{2x} + \frac{2}{x^3} + 2 \ln(3) \cdot 3^{-2x} + 9 \]