Question: 4 Find \( f^{\prime}(x) \) for the function \( f(x)=\frac{\tan x+\csc ^{-1} x-3 \log _{5} x}{\tan ^{-1} x+\sec x-2^{x}} \)

Finding \( f^{\prime}(x) \) for the given function \( f(x)=\frac{\tan x+\csc^{-1} x-3 \log_{5} x}{\tan^{-1} x+\sec x-2^{x}} \) involves applying the quotient rule. The quotient rule states that if you have a function in the form of \( \frac{u(x)}{v(x)} \), then the derivative \( f'(x) \) can be calculated as: \[ f'(x) = \frac{u'v - uv'}{v^2} \] In this case, you would define: - \( u(x) = \tan x + \csc^{-1} x - 3 \log_{5} x \) - \( v(x) = \tan^{-1} x + \sec x - 2^{x} \) You will need to compute \( u'(x) \) and \( v'(x) \): - The derivative of \( u(x) \) will involve the derivatives of \( \tan x \), \( \csc^{-1} x \), and \( \log_{5} x \). - The derivative of \( v(x) \) will involve the derivatives of \( \tan^{-1} x \), \( \sec x \), and \( 2^{x} \). Once you have \( u' \) and \( v' \), plug those into the quotient formula above to arrive at \( f'(x) \). Tip: Always remember to simplify your final answer! It helps in ensuring that it’s easy to read and gives a clearer understanding of the behavior of the function. Don't forget to apply the chain rule where necessary, especially for compositions of functions. When dealing with intricate functions like this, it can be easy to overlook negative signs or coefficients when calculating derivatives. Double-check each step to avoid tripping up on a simple mistake!