Respuesta :
Answer:
[tex]f'(x)=-8\sin2x+\frac{1}{x} +1[/tex]
Step-by-step explanation:
I assume by [tex]\log(x)[/tex], the base is [tex]e[/tex], so let [tex]\log_e(x)=\ln(x)[/tex]:
[tex]\displaystyle f(x)=4\cos2x+\ln x+x\\\\f'(x)=\frac{d}{dx}(4\cos2x+\ln x + x)\\\\f'(x)=\frac{d}{dx}(4\cos2x)+\frac{d}{dx}(\ln x)+\frac{d}{dx}(x)\\\\f'(x)=-8\sin2x+\frac{1}{x}+1[/tex]
Remember that by the chain rule, to find the derivative of [tex]4\cos2x[/tex], you must find the derivative of the inside, which is [tex]\displaystyle \frac{d}{dx}(2x)=2[/tex], and multiply it by the outside derivative. Recall that [tex]\displaystyle \frac{d}{dx}\cos(x)=-\sin(x)[/tex], so replace [tex]-\sin(x)[/tex] with [tex]-4\sin(2x)[/tex], and multiply by 2 to get [tex]-8\sin(2x)[/tex] as your derivative.
Also remember that [tex]\displaystyle \frac{d}{dx}\ln(x)=\frac{1}{x}[/tex], one of the more basic derivatives.
