Given y=1/3cosx,y'= 1/3sinx -3sinx -1/3sinx 3cosx

To find the derivative of the function $y = \frac{1}{3} \cos x$, we can use the chain rule.<br /><br />The chain rule states that if we have a composite function $y = f(g(x))$, then the derivative of $y$ with respect to $x$ is given by:<br /><br />$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$<br /><br />In this case, we have $y = \frac{1}{3} \cos x$, which can be written as $y = f(x) = \frac{1}{3} \cos x$, where $f(x) = \frac{1}{3} \cos x$.<br /><br />Now, let's find the derivative of $f(x)$ with respect to $x$:<br /><br />$f'(x) = \frac{d}{dx} \left( \frac{1}{3} \cos x \right) = \frac{1}{3} \cdot \frac{d}{dx} (\cos x) = \frac{1}{3} \cdot (-\sin x) = -\frac{1}{3} \sin x$<br /><br />Therefore, the derivative of $y = \frac{1}{3} \cos x$ is:<br /><br />$y' = f'(x) = -\frac{1}{3} \sin x$<br /><br />So, the correct answer is $-\frac{1}{3} \sin x$.