Question

Find the intervals of increasing and decreasing of the function

$f(x) = \frac{4 \: sinx}{2 + cosx} - x \: \: on \: (0 \: to \: \frac{\pi}{2})$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = \dfrac{4sinx}{2 + cosx} - x$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx} f(x) = \dfrac{d}{dx}\dfrac{4sinx}{2 + cosx} - \dfrac{d}{dx}x$

We know,

$\boxed{\tt{ \dfrac{d}{dx} \frac{u}{v} \: = \: \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: }}$

and

$\boxed{\tt{ \dfrac{d}{dx}x= 1}}$

So, using these results, we get

$\rm \: f'(x) = \dfrac{(2 + cosx)\dfrac{d}{dx}(4sinx) - 4sinx\dfrac{d}{dx}(2 + cosx)}{ {(2 + cosx)}^{2} } - 1$

We know,

$\boxed{\tt{ \dfrac{d}{dx}sinx = cosx}}$

and

$\boxed{\tt{ \dfrac{d}{dx}cosx = - \: sinx}}$

So, using these results, we get

$\rm \: f'(x) = \dfrac{(2 + cosx)(4cosx) - (4sinx)(0 - sinx)}{ {(2 + cosx)}^{2} } - 1$

$\rm \: f'(x) = \dfrac{8cosx + {4cos}^{2}x + {4sin}^{2}x}{ {(2 + cosx)}^{2} } - 1$

$\rm \: f'(x) = \dfrac{8cosx +4( {cos}^{2}x + {sin}^{2}x)}{ {(2 + cosx)}^{2} } - 1$

$\rm \: f'(x) = \dfrac{8cosx +4( 1)}{ {(2 + cosx)}^{2} } - 1$

$\rm \: f'(x) = \dfrac{8cosx +4}{ {(2 + cosx)}^{2} } - 1$

$\rm \: f'(x) = \dfrac{8cosx +4 - {(2 + cosx)}^{2} }{ {(2 + cosx)}^{2} }$

$\rm \: f'(x) = \dfrac{8cosx +4 - (4 + {cos}^{2}x + 4cosx) }{ {(2 + cosx)}^{2}}$

$\rm \: f'(x) = \dfrac{8cosx +4 - 4 - {cos}^{2}x - 4cosx}{ {(2 + cosx)}^{2}}$

$\rm \: f'(x) = \dfrac{4cosx - {cos}^{2}x}{ {(2 + cosx)}^{2}}$

$\rm \: f'(x) = \dfrac{cosx(4 - {cos}x)}{ {(2 + cosx)}^{2}}$

As it is given that,

$\rm \: x \: \in \: \bigg(0, \: \dfrac{\pi}{2}\bigg)$

$\rm\implies \:cosx \: \in \: (0,1)$

and

$\rm \: 4 - cosx > 0 \: \: and \: \: {(2 + cosx)}^{2} > 0$

$\rm\implies \:f'(x) > 0$

$\rm\implies \:f'(x) \: is \: strictly \: increasing \: on \: \bigg(0, \: \dfrac{\pi}{2} \bigg)$

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MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

So,we have the function f(x) and asked to check the interval where it is increasing and decreasing. So the very basic thing which could be done is to differentiate f(x) and there on apply the definition of range,domain,etc whatever is required as per the given function.

=> f'(x) = {4 [(2+cosx) (cos) - (sinx) (0 - sinx)]/(2+cosx)²} - 1

[Took 4 outside because it is a constant and we know that : d[(k. f(x)]/dy = k. f'(x) where k is a constant]

=> f'(x) = {4[(2cosx + cos²x - (sinx) (-sinx)]/(2+cosx)²} - 1

=> f'(x) = {4[2cosx + cos²x + sin²x)]/(2+cosx)²} - 1

=> f'(x) = [4(2cosx + 1)/(2+cosx)²] - 1...[co²x + sin²x =1)

=> f'(x) = [(8cosx + 4)/ (2+cosx)²] - 1

=> f'(x) = [8cosx + 4 - (2+cosx)²]/(2+cosx)²...(taking LCM)

=> f'(x) = [8cosx + 4 -(4 + cos²x + 4cosx)]/(2+cosx)²

=> f'(x) = (8cosx + 4 - 4 - cos²x - 4cosx)/(2+cosx)²

=> f'(x) = (8cosx - cos²x - 4cosx)/(2+cosx)²

=> f'(x) = (4cosx - cos²x)/(2+cosx)²

=> f'(x) = [cosx (4 - cosx)/(2+cosx)²

The denominator of f'(x) is always going to be positive,so we need not check it.

Let's check the numerator.

=> cosx (4- cosx) > 0

The term which you can see in the bracket will also be always positive,why? Because range of cosx is from -1 to 1, but maybe you might wonder that if you put cosx = - 1 then we would be left with cos(-5) which looks like something negaaaative...? Well, reminding you that cos isn't really a very decent guy like me xD if you input any negative angle in it,it will eat your negative sign xD moreover look at your interval you gave in the question,it aint gonna allow the negative values. So,

=> cos x > 0

And this is always true in your mentioned interval,i.e (0 to π/2).

So,here since the numerator and denominator both are very positive ppl, so your f(x) is a strictly increasing function on (0 to π/2).

(not really having good LaTeX skills,so sorry for the mess xD but being a 11th/12th grader you probably shouldn't be complaining about the excess of brackets xD I hate it too though but can't help)