Question

Hlo guyz..Needed step by step xplanation...
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Answer 1

QUESTION:

$\sf{If \ {y = \dfrac{sin \ x}{1 + \dfrac{cos \ x}{1 + \dfrac{sin \ x}{1 + \dfrac{sin \ x}{1 + \dots \ \dots \ \dots + \infty} } }} }}$

$\sf{{prove \ that \ \dfrac{dy}{dx}= \dfrac{(1+y)cos \ x}{1+2y+cos \ x-sin\ x}}}$

GIVEN:

$\sf{ {y = \dfrac{sin \ x}{1 + \dfrac{cos \ x}{1 + \dfrac{sin \ x}{1 + \dfrac{sin \ x}{1 + \dots \ \dots \ \dots + \infty} } }} }}$

TO PROVE:

$\sf {\dfrac{dy}{dx}= \dfrac{(1+y)cos \ x}{1+2y+cos \ x-sin\ x}}$

EXPLANATION:

$\tt{{y = \dfrac{sin \ x}{1 + \dfrac{cos \ x}{1 + \dfrac{sin \ x}{1 + \dfrac{sin \ x}{1 + \dots \ \dots \ \dots + \infty} } } }}}$

$\tt{{y = \dfrac{sin \ x}{1 + \dfrac{cos \ x}{1 +y} } }}$

$\tt{{y = \dfrac{sin \ x}{\dfrac{1 + y + cos \ x}{1 +y} }} }$

$\tt{{y = \dfrac{sin \ x(1 + y)}{1 + y + cos \ x} }}$

$\tt{{y + {y}^{2} + ycos \ x = sin \ x +ysin \ x }}$

Let us split the terms.

$\tt{Take \ y + {y}^{2} + ycos \ x }$

Differentiate w.r.t x

$\red{\bigstar\boxed{ \bold{ \gray{ \dfrac{d}{dx} {x}^{2} = 2x}}}}$

$\blue{\bigstar\boxed{ \bold{ \gray{\dfrac{d}{dx} uv = uv' + vu'}}}}$

$\orange{\bigstar\boxed{ \bold{ \gray{\dfrac{d}{dx}cos \ x = - sin \ x}}}}$

$\tt{ {\dfrac{dy}{dx} + 2y \dfrac{dy}{dx} + \dfrac{dy}{dx} cos \ x + y \dfrac{d}{dx} cos \ x}}$

$\tt{\dfrac{dy}{dx}(1 + 2y + cos \ x) - y sin \ x}$

$\tt{Now \ take \ = sin \ x +ysin \ x}$

Differentiate w.r.t x

$\purple{\bigstar\boxed{ \bold{ \gray{\dfrac{d}{dx} sin \ x = cos \ x}}}}$

$\pink{\bigstar\boxed{ \bold{ \gray{\dfrac{d}{dx} uv = uv' + vu'}}}}$

$\tt{\dfrac{d}{dx} sin \ x +y \dfrac{d}{dx} sin \ x + sin \ x\dfrac{dy}{dx}}$

$\tt{cos \ x +y cos \ x + sin \ x\dfrac{dy}{dx}}$

Now equate the terms

dy/dx(1 + 2y + cos x) - y sin x = cos x + y cos x + sin x(dy/dx)

dy/dx(1 + 2y + cos x) - sin x(dy/dx) = cos x (1 + y ) + y sin x

dy/dx(1 + 2y + cos x - sin x) = cos x (1 + y) + y sin x

$\tt{\dfrac{dy}{dx}= \dfrac{(1+y)cos \ x}{1+2y+cos \ x-sin\ x}}$

HENCE PROVED.