Show that $f(x)=\dfrac{\sin x}{x}$ is monotonically decreasing on $[0,\frac{\pi}{2}]$
I'm trying to show that $f'(x)\leq0$ to show it's monotonically decreasing. So $f'(x)=\dfrac{x\cos x-\sin x}{x^2}$. I can see that $f'(x)\leq0$ but having trouble proving it.
We have using the formula for the derivative of a quotient $$ \frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin(x)}{x}=\frac{x\cos(x)-\sin(x)}{x^2} $$ Then we can use the fact that on $[0,\pi/2]$ $$ \tan(x)\ge x\qquad\text{and}\qquad\cos(x)\ge0 $$ See $(1)$ or $(4)$ of this answer.
To show $f'(x)\lt 0$ in the interval $(0,\pi/2)$, it is enough to show that $g(x)\lt 0$ in the interval, where $g(x)=x\cos x-\sin x$.
Note that $g(0)=0$ and $g'(x)=-x\sin x$. Thus $g(x)$ is decreasing in the interval $(0,\pi/2)$, indeed all the way to $\pi$. It follows that $g(x)\lt 0$ in our interval.
