Question

if u = x/2 , then find derivative ( 2sinx - sin2x) / (2sinx + sin2x)

Answer 1

Solution :

$\sf{First\:let\:us\:solve\:the\:identity\::\:\dfrac{2sin(x) - sin2(x)}{2sin(x) + sin2(x)}}$

$:\implies \sf{\dfrac{2sin(x) - sin2(x)}{2sin(x) + sin2(x)}}$

$:\implies \sf{\dfrac{2sin(x) - 2sin(x)cos(x)}{2sin(x) + 2sin(x)cos(x)}} \:\:\:[\because \sf{sin2\theta = 2sin(\theta)cos(\theta)}]$

$:\implies \sf{\dfrac{2sin(x)[1 - cos(x)]}{2sin(x)[1 + cos(x)]}}$

$:\implies \sf{\dfrac{1 - cos(x)}{1 + cos(x)}}$

$\textsf{Now, by using the identity of cos(x)\:} \\ \textsf{and substituting them in the equation, we get :} \\ \\ \bullet\:\sf{cos(x) = 2cos^{2}\bigg(\dfrac{x}{2} - 1\bigg)} \\ \\ \bullet\:\sf{cos(x) = 1 - 2sin^{2}\bigg(\dfrac{x}{2}\bigg)}$

$:\implies \sf{\dfrac{1 - 1 - 2sin^{2}\bigg(\dfrac{x}{2}\bigg)}{1 + 2cos^{2}\bigg(\dfrac{x}{2}\bigg) - 1}}$

$:\implies \sf{\dfrac{2sin^{2}\bigg(\dfrac{x}{2}\bigg)}{2cos^{2}\bigg(\dfrac{x}{2}\bigg)}}$

$\textsf{Now, by using the identity of cos(x)\:} \\ \textsf{and substituting them in the equation, we get :} \\ \mapsto \sf{\dfrac{sin(x)}{cos(x)} = tan(x)}$

$:\implies \sf{tan^{2}\bigg(\dfrac{x}{2}\bigg)}$

$\boxed{\therefore \sf{\dfrac{2sin(x) - sin2(x)}{2sin(x) + sin2(x)} = tan^{2}\bigg(\dfrac{x}{2}\bigg)}}$

$\textsf{We know that the value of\:}\sf{u\:is\: \dfrac{x}{2}}\:\textsf{So by substituting it in the equation, we get :}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$:\implies \sf{tan^{2}\bigg(\dfrac{x}{2}\bigg)}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$:\implies \sf{tan^{2}\bigg(\dfrac{x}{2}\bigg) = tan^{2}(u)}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\boxed{\therefore \sf{tan^{2}\bigg(\dfrac{x}{2}\bigg) = tan^{2}(u)}}$

$\textsf{Now, by differentiating the function with respect to u, we get :-}$

$:\implies \sf{\dfrac{dy}{du} = \dfrac{d[tan^{2}u]}{du}}$

$\textsf{Now, by applying the chain rule of differentiation, we get :-} \\ \\ \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

$:\implies \sf{\dfrac{dy}{du} = \dfrac{d[tan^{2}(u)}{d[tan(u)]}\cdot\dfrac{d[tan(u)]}{du}}$

$:\implies \sf{\dfrac{dy}{du} = 2tan(u)\cdot sec^{2}(u)}$

$:\implies \sf{\dfrac{dy}{du} = 2tan(u)sec^{2}(u)}$

$\boxed{\therefore \sf{\dfrac{dy}{du} = 2tan(u)(sec^{2}(u)}}$