Question

homework4: problem 2
(1 point)
let ( f(x)=\frac{\tan (x)-3}{sec (x)} ). find the following:

1. ( f^{prime}(x)=cos (x)+sin (x) )
2. ( f^{prime}(3)=cos (3)+sin (x) )

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Explanation:

Step1: Simplify 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 \((\sin(x))^\prime=\cos(x)\) and \((\cos(x))^\prime =-\sin(x)\), for \(y = f(x)=\sin(x)-3\cos(x)\), then \(f^\prime(x)=\cos(x)+ 3\sin(x)\)

Step3: Evaluate \(f^\prime(3)\)

Substitute \(x = 3\) into \(f^\prime(x)\), we get \(f^\prime(3)=\cos(3)+3\sin(3)\)

Answer:

1. \(f^\prime(x)=\cos(x)+3\sin(x)\)
2. \(f^\prime(3)=\cos(3)+3\sin(3)\)