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

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

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

Answer

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)