Q.Which of the following function is strictly decreasing function in interval $\left(0,\dfrac{\pi}{2}\right)$?

$f$ is decreasing exactly where $f'(x)<0$ AND the function is actually defined there (needs $\cos x>0$).

$f(x) = \log(\cos x)$, so $f'(x) = \dfrac{-\sin x}{\cos x} = -\tan x$.

For $f$ to be strictly decreasing, we need $f'(x) < 0 \Rightarrow \tan x > 0$.

But we also need $\cos x > 0$ for $\log(\cos x)$ to even be defined (real-valued).

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A function is strictly decreasing on an interval where its derivative is negative there.

On $\left(0,\dfrac{\pi}{2}\right)$: $\dfrac{d}{dx}(\sin x)=\cos x>0$ (increasing); $\dfrac{d}{dx}(\cos x)=-\sin x<0$ (decreasing); $\dfrac{d}{dx}(\tan x)=\sec^2 x>0$ (increasing); $\dfrac{d}{dx}(\sin 2x)=2\cos 2x$, which changes sign in this interval (not strictly monotonic throughout).

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$\sin x$ increases exactly where $\cos x>0$; among the options that is $(0,\pi/2)$.

Let $f(x)=\sin x$. Then $f'(x)=\cos x$.

$f$ is increasing on an interval where $f'(x)=\cos x>0$.

On $(0,\pi/2)$: $\cos x>0$ throughout, so $f$ is increasing here.

On $(\pi/2,\pi)$: $\cos x<0$, so $f$ is decreasing.

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$\sin x$ increases exactly where $\cos x>0$; check each given interval against that sign.

$f(x)=\sin x \Rightarrow f'(x)=\cos x$. $f$ is increasing on an interval where $\cos x>0$.

• $(\pi/2,\pi)$: $\cos x<0$ here (second quadrant) — decreasing, not this one.
• $(\pi,3\pi/2)$: $\cos x<0$ here (third quadrant) — decreasing, not this one. …

From the graph, $\sin x$ rises on $(0,\pi/2)$ and falls on $(\pi/2,\pi)$, so it is not monotonic on the whole interval $(0,\pi)$.

From the given graph of $f(x) = \sin x$: on $(0, \pi/2)$ the curve rises from $0$ to its maximum value $1$ at $x=\pi/2$, so $f'(x) = \cos x > 0$ there and $f$ is increasing on $(0,\pi/2)$.

On $(\pi/2, \pi)$ the curve falls from $1$ back down to $0$ at $x=\pi$, so $f'(x) = \cos x < 0$ there and $f$ is decreasing on $(\pi/2,\pi)$.

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Since $f'(x)=\cos x - 1 \le 0$ for every real $x$, the function never increases; it is monotonically decreasing throughout $\mathbb{R}$.

Differentiate: $f'(x) = \cos x - 1$.

Since $\cos x \le 1$ for all real $x$, we get $f'(x) = \cos x - 1 \le 0$ for every $x \in \mathbb{R}$, with equality only at the isolated points $x = 2n\pi$, $n \in \mathbb{Z}$.

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$f'(x)=\cot x$, which is negative on $\left(\frac{\pi}{2},\pi\right)$, making $f$ decreasing there.

$f'(x) = \dfrac{\cos x}{\sin x} = \cot x$

On $\left(0,\dfrac{\pi}{2}\right)$, $\cot x>0$ so $f$ is increasing. …