Question

let ( f(x)=c x+ln (cos (x)) ). for what value of ( c ) is ( f^{prime}left(\frac{pi}{4}
ight)=2 )?

Explanation:

Step1: Differentiate \(f(x)\)

Using the sum rule \((u + v)^\prime=u^\prime+v^\prime\) and the chain - rule \((\ln(u))^\prime=\frac{u^\prime}{u}\).
If \(u = \cos(x)\), then \(u^\prime=-\sin(x)\).
\(f(x)=cx+\ln(\cos(x))\), so \(f^\prime(x)=c+\frac{-\sin(x)}{\cos(x)}=c - \tan(x)\)

Step2: Substitute \(x = \frac{\pi}{4}\) into \(f^\prime(x)\)

We know that \(\tan(\frac{\pi}{4}) = 1\).
Since \(f^\prime(\frac{\pi}{4})=2\), substitute \(x=\frac{\pi}{4}\) into \(f^\prime(x)\):
\(f^\prime(\frac{\pi}{4})=c-\tan(\frac{\pi}{4})\)
\(2=c - 1\)

Step3: Solve for \(c\)

Add \(1\) to both sides of the equation \(2=c - 1\).
\(c=2 + 1\)

Answer:

\(3\)