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if ( f ( x ) = \frac { \tan x - 3 } { sec x } )
( f ^ { prime } ( x ) = )
( f ^ { prime } ( 1 ) = )
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Explanation:

Step1: Rewrite the function

We know that $\tan x=\frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. So $f(x)=\frac{\tan x - 3}{\sec x}=\sin x-3\cos x$.

Step2: Differentiate the function

Using the derivative rules: $( \sin x)'=\cos x$ and $( \cos x)'=-\sin x$.
The derivative of $y = f(x)=\sin x-3\cos x$ is $f'(x)=\cos x+3\sin x$.
Since $\cos x=\frac{1}{\sec x}$ and $\sin x=\frac{\tan x}{\sec x}$, we can also write $f'(x)=\sec x - 3\tan x\sec x$.

Step3: Evaluate the derivative at \(x = 1\)

Substitute \(x = 1\) into \(f'(x)\).
\(f'(1)=\cos(1)+3\sin(1)\), or using the \(\sec\) and \(\tan\) form \(f'(1)=\sec(1)-3\tan(1)\sec(1)\) (where the angle \(x = 1\) is in radians).

Answer:

$f'(x)=\sec x - 3\tan x\sec x$; $f'(1)=\sec(1)-3\tan(1)\sec(1)$