Which of the following is not a decreasing fu | vedclass

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(C) function $f(x)$ is decreasing on an interval if $f'(x) \leq 0$ for all $x$ in that interval.$(A)$ For $f(x) = \cos x$,$f'(x) = -\sin x$. In $\left

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$\cos 3x$

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(C) function $f(x)$ is decreasing on an interval if $f'(x) \leq 0$ for all $x$ in that interval.$(A)$ For $f(x) = \cos x$,$f'(x) = -\sin x$. In $\left( 0, \frac{\pi}{2} \right)$,$\sin x > 0$,so $f'(x) $(B)$ For $f(x) = \cos 2x$,$f'(x) = -2\sin 2x$. In $\left( 0, \frac{\pi}{2} \right)$,$2x \in (0, \pi)$,where $\sin 2x > 0$,so $f'(x) $(C)$ For $f(x) = \cos 3x$,$f'(x) = -3\sin 3x$. In $\left( 0, \frac{\pi}{2} \right)$,$3x$ ranges from $0$ to $\frac{3\pi}{2}$. In the interval $\left( \frac{\pi}{3}, \frac{\pi}{2} \right)$,$3x$ is in $\left( \pi, \frac{3\pi}{2} \right)$,where $\sin 3x  0$. Therefore,$\cos 3x$ is not decreasing on the entire interval.$(D)$ For $f(x) = \cot x$,$f'(x) = -\csc^2 x$. Since $\csc^2 x > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$,$f'(x) < 0$. Thus,$\cot x$ is decreasing.

Which of the following is not a decreasing function on the interval $\left( 0, \frac{\pi}{2} \right)$?

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