Question

what is the derivative of 4 cos(x) - ln(x)?
4 sin(x) - x ln(x)
sin(4x) - \frac{1}{x}

• sin(4x) - \frac{1}{x}

-4 sin(x) - \frac{1}{x}
-4 sin(x) - x ln(x)
4 sin(x) - \frac{1}{x}
clear my selection

Explanation:

Step1: Differentiate \(4\cos(x)\)

Using the rule \(\frac{d}{dx}(a\cos(x))=-a\sin(x)\), for \(a = 4\), we have \(\frac{d}{dx}(4\cos(x))=-4\sin(x)\)

Step2: Differentiate \(-\ln(x)\)

Using the rule \(\frac{d}{dx}(\ln(x))=\frac{1}{x}\), so \(\frac{d}{dx}(-\ln(x))=-\frac{1}{x}\)

Step3: Combine the derivatives

By the sum - rule of differentiation \(\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}\), if \(u = 4\cos(x)\) and \(v=-\ln(x)\), then \(\frac{d}{dx}(4\cos(x)-\ln(x))=-4\sin(x)-\frac{1}{x}\)

Answer:

\(-4\sin(x)-\frac{1}{x}\) (the fourth option)