Question

y=sin^-1(4+5sinx/5+4 sin x) then find dy/dx

Answer 1

Step-by-step explanation:

We have,

$y = \sin^{ - 1} \bigg( \frac{4 + 5 \sin(x) }{5 + 4 \sin(x) } \bigg)$

$\implies \frac{dy}{dx}= \frac{1}{ \sqrt{1 - \bigg( \frac{4 + 5 \sin(x) }{5 + 4 \sin(x) } \bigg)^{2} }}. \frac{d}{dx} \bigg \{ \frac{4 + 5 \sin(x) }{5 + 4 \sin(x) } \bigg \}$

$\implies \frac{dy}{dx}= \frac{5 + 4 \sin(x) }{ \sqrt{(5 + 4 \sin(x))^{2} - ( 4 + 5 \sin(x) )^{2} }}. \bigg \{ \frac{ 5 \cos(x) (5 + 4 \sin(x)) - 4 \cos(x)(4 + 5 \sin(x) )}{(5 + 4 \sin(x))^{2} } \bigg \}$

$\implies \frac{dy}{dx}= \frac{1 }{ \sqrt{25 + 16\sin^{2} (x) + 40 \sin(x) \cos(x) - 16 - 25 \sin^{2} (x) - 40 \sin(x) \cos(x) }}. \bigg \{ \frac{ 25 \cos(x) + 20\sin(x) \cos(x) - 16 \cos(x) - 20 \sin(x) \cos(x) }{5 + 4 \sin(x) } \bigg \}$

$\implies \frac{dy}{dx}= \frac{1 }{ \sqrt{9 - 9\sin^{2} (x) }}. \bigg \{ \frac{ 9 \cos(x) }{5 + 4 \sin(x) } \bigg \}$

$\implies \frac{dy}{dx}= \frac{1 }{ \sqrt{9(1 - \sin^{2} (x) ) }}. \bigg \{ \frac{ 9 \cos(x) }{5 + 4 \sin(x) } \bigg \}$

$\implies \frac{dy}{dx}= \frac{1 }{ 3 \cos (x) }. \bigg \{ \frac{ 9 \cos(x) }{5 + 4 \sin(x) } \bigg \}$

$\implies \frac{dy}{dx}= \frac{ 3 }{5 + 4 \sin(x) }$