Question

If y=log(1+cos x), prove that y3+y2y1=0​

Answer 1

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

Given that,

$\rm :\longmapsto\:y = log(1 + cosx)$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} log(1 + cosx)$

We know,

$\boxed{ \rm \:\dfrac{d}{dx}logx = \frac{1}{x}}$

So, using this, we get

$\rm :\longmapsto\:y_1 = \dfrac{1}{1 + cosx}\dfrac{d}{dx}(1 + cosx)$

We know,

$\boxed{ \rm \:\dfrac{d}{dx}k = 0}$

and

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

So, using this,

$\rm :\longmapsto\:y_1 = \dfrac{1}{1 + cosx}(0 - sinx)$

$\rm :\longmapsto\:y_1 = - \: \dfrac{sinx}{1 + cosx}$

can be further reduced as

$\rm :\longmapsto\:y_1 = - \: \dfrac{2sin\bigg[\dfrac{x}{2} \bigg]cos\bigg[\dfrac{x}{2} \bigg]}{2 {cos}^{2} \bigg[\dfrac{x}{2} \bigg]}$

$\rm :\longmapsto\:y_1 = - \: tan\dfrac{x}{2} - - - - (1)$

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

$\rm :\longmapsto\:\dfrac{d}{dx} \: y_1 = - \:\dfrac{d}{dx} \: tan\dfrac{x}{2}$

We know,

$\boxed{ \rm \:\dfrac{d}{dx}tanx = {sec}^{2}x}$

So, using this, we get

$\rm :\longmapsto\:y_2 = - \: {sec}^{2} \bigg[\dfrac{x}{2} \bigg]\dfrac{d}{dx}\dfrac{x}{2}$

$\rm :\longmapsto\:y_2 = - \: {sec}^{2} \bigg[\dfrac{x}{2} \bigg] \times \dfrac{1}{2}$

$\rm :\longmapsto\:y_2 = - \:\bigg(\dfrac{1}{2} \bigg) {sec}^{2} \bigg[\dfrac{x}{2} \bigg]$

can be rewritten as

$\rm :\longmapsto\:y_2 = - \dfrac{1}{2} \bigg[1 + \: {tan}^{2} \bigg[\dfrac{x}{2} \bigg]\bigg]$

$\rm :\longmapsto\:y_2 = - \dfrac{1}{2} (1 + {y_1}^{2}) \: \: \: \: \: \: \: \: \{ \: using \: (1) \: \}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y_2 = - \: \dfrac{1}{2} \dfrac{d}{dx}(1 + {y_1}^{2})$

$\rm :\longmapsto\:y_ 3 \: = \: - \dfrac{1}{2} \bigg(0 + 2y_1\dfrac{d}{dx}y_1\bigg)$

$\rm :\longmapsto\:y_ 3 \: = \: - y_1 \: y_2$

$\bf\implies \:y_3 + y_1 \: y_2 = 0$

Hence, Proved

Additional Information :-

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