Question

Y=sin(2sin^-1x) show that dy/dx=2root of 1 -y^2/root of 1 -x^2

Answer 1

Step-by-step explanation:

$y = sin(2 {sin}^{ - 1} x) \\ \frac{dy}{dx} = \frac{d}{dx} sin(2 {sin}^{ - 1} x )\\ = cos \: (2 {sin}^{ - 1} x) \times \frac{d}{dx} (2 {sin}^{ - 1} x) \\ = cos \: (2 {sin}^{ - 1} x) \times \frac{2}{ \sqrt{1 - {x}^{2} } } \\ = \frac{2 cos \: (2 {sin }^{ - 1} x)}{ \sqrt{1 - {x}^{2} } } \: \\ = \frac{2 \sqrt{1 - {sin}^{2}(2 {sin}^{ - 1} x} )}{ \sqrt{1 - {x}^{2} } } \\ = \frac{2 \sqrt{1 - {y}^{2}}}{ \sqrt{1 - {x}^{2} } }$