Question

Let f(x) = cx + ln(cos(x)). for what value of c is f '(π/4) = 6?

Answer 1

let f(x) = cx + ln(cos(x)).

Take the derivative of the function.

f(x) = cx + ln(cos(x)

f'(x) =c + 1/cos(x)*-sin(x)

f'(x) =c-tan(x)

we take the derivative of the outside times the derivative of inside,by using the chain rule.

so,to find the derivative of  ln(cos(x), first we take the derivative of ln(u) where u=cos(x) that is 1/u or 1/cos(x)

now,multiply the derivative of u or cos(x) that is sin(x).By combining and simplfying the derivatives

the equation will be;

f'(x)=c-tan(x)

In that case

we are finding c when  f '(π/4) = 5

equation become;

f'(π/4)=c-tan(π/4)=5

where tan(π/4)=1

putting the values  

c-1=5

c=5+1

so, c=6

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