Question

If sin y = x sin ( a + y ) , prove that dy = sin " ( 0 + 3 ) dx _____
sin a​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: If \: siny = x \: sin(a + y) \: Prove that \: \dfrac{dy}{dx} = \dfrac{{sin}^{2} (a + y)}{sina}$

$\large\underline{\bold{Answer-}}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$(1). \: \: \boxed{ \bf{ \: \dfrac{d}{dx}sinx = cosx}}$

$(2). \: \: \boxed{ \bf{ \: \dfrac{d}{dx}k = 0}}$

$(3). \: \: \boxed{ \bf{ \: \dfrac{d}{dx}x = 1}}$

$(4). \: \: \boxed{ \bf{sin(x - y) = sinxcosy + sinycosx}}$

$(5) \: \: \boxed{ \bf{ \: \dfrac{d}{dx}\bigg( \dfrac{u}{v} \bigg) = \dfrac{v \: \dfrac{d}{dx}u \: - \: u \: \dfrac{d}{dx}v}{ {v}^{2} } }}$

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

$\bf \: siny = x \: sin(a + y)$

$\sf \: x = \dfrac{siny}{sin(a + y)}$

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

$\sf \: \dfrac{d}{dy}x = \dfrac{d}{dy}\bigg(\dfrac{siny}{sin(a + y)} \bigg)$

$\sf \: \dfrac{dx}{dy} = \dfrac{sin(a + y)\dfrac{d}{dy}siny - siny\dfrac{d}{dy}sin(a + y)}{ {sin}^{2} (a + y)}$

$\sf \: \dfrac{dx}{dy} = \dfrac{sin(a + y) \: cosy \: - \: siny \: cos(a + y)}{ {sin}^{2}(a + y) }$

$\sf \: \dfrac{dx}{dy} = \dfrac{sin(a + \cancel{y} - \cancel{y})}{ {sin}^{2}(a + y)}$

$\therefore \: \bf \: \dfrac{dy}{dx} \: = \: \dfrac{ {sin}^{2} (a + y)}{sina}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}$

$\boxed{ \bf{ \: \dfrac{d}{dx} log(x) = \dfrac{1}{x} }}$

$\boxed{ \bf{ \: \dfrac{d}{dx} {e}^{x} = {e}^{x} }}$

$\boxed{ \bf{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) \: \: \: where \: a > 0}}$

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

$\boxed{ \bf{ \: \dfrac{d}{dx}secx = secx \: tanx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}cotx = - {cosec}^{2} x}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}cosecx = - cosecx \: cotx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}u.v \: = \: u \: \dfrac{d}{dx} v \: + \: v \: \dfrac{d}{dx} u}}$