The differentiation of the function y = sin(2 sin⁻¹x) is; dy/dx = d√(1 - y²)]/(√1 - x²)
How to differentiate functions?
1) We want to differentiate the function;
[tex]\frac{\sqrt{x + 1} + \sqrt{x - 1}}{\sqrt{x + 1} - \sqrt{x - 1} }[/tex]
Differentiating this gives;
dy/dx = [tex]\frac{[(\frac{1}{2\sqrt{x + 1}} - \frac{1}{2\sqrt{x - 1}}) * \sqrt{x + 1} + \sqrt{x - 1}]}{(\sqrt{x + 1} - \sqrt{x - 1})^{2} }[/tex]
That can be simplified to get;
dy/dx = [tex]\frac{\sqrt{x + 1} + \sqrt{x - 1} }{(x - 1)\sqrt{x + 1} + (-x - 1) \sqrt{x - 1}}[/tex]
2) We want to differentiate the function;
y = sin(2 sin⁻¹x)
Differentiating the function gives;
dy/dx = [2cos(2 sin⁻¹x)]/√(1 - x²)
We know from trigonometric identity that;
cos²x = 1 - sin²x
Thus;
1 - sin²(2 sin⁻¹x) = cos²(2 sin⁻¹x)
But sin²(2 sin⁻¹x) = y²cos²(2 sin⁻¹x)
Thus; √(1 - y²) = cos(2 sin⁻¹x)
dy/dx = d√(1 - y²)]/(√1 - x²)
Read more about Differentiation of Functions at; https://brainly.com/question/5313449
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