Question

Hlo...solve stepwise . .​

Answer 1

Answer:

your answer attached in the photo

Answer 2

Above answer is nicely solved,but u can also proceed through this method to simiplify the term which needs to be diffrentiated

As we know

${sin}^{ - 1} x 1+ {sin}^{ - 1} y1= {sin}^{ - 1} (x1 \sqrt{1 - {(y1)}^{2} } + y1 \sqrt{1 - {(x1)}^{2} } \\ compare \: the \: terms \: with \: \\ above \: rhs \\ {sin}^{ - 1} (x1 \sqrt{1 - {(y1)}^{2} } + y1 \sqrt{1 - {(x1)}^{2} } = {sin}^{ - 1} ( {x}^{2} \sqrt{1 - {x}^{2} } + x \sqrt{1 - {x}^{4} }) \\ comparing \: both \: side \: we \: get x1 = {x}^{2 } \\ and \: y1 = x \\ so \: we \: can \: write \\ y = {sin}^{ - 1} {x}^{2} + {sin}^{ - 1}x = {sin}^{ - 1} ( {x}^{2} \sqrt{1 - {x}^{2} } + x \sqrt{1 - {x}^{4} } \\ now \: we \: know \\ {sin}^{ - 1} x = \frac{1}{ \sqrt{1 - {x}^{2} } } \\ so \: \\ \frac{dy}{dx} = \frac{1}{ \sqrt{1 - {x}^{4} } }.2x + \frac{1}{ \sqrt{1 - {x}^{2} } } \\ \frac{dy}{dx} = \frac{2x}{ \sqrt{1 - {x}^{4} } } + \frac{1}{ \sqrt{1 - {x}^{2} } }$