Question

What is the derivative of y=tan(arcsin(x))?

Answer 1

Hello,

(tg(x))'=1/cos²(x)

(arcsin(x))'=1/√(1-x²)

cos²(arcsin(x))=1-sin²(arcsin(x))=1-x²


$(tg(arcsin(x)))'= \dfrac{1}{cos(arcsin(x)))^2} * \dfrac{1}{\sqrt{1-x^2}} \\ =\dfrac{1}{1-sin(arcsin(x))^2}* \dfrac{1}{\sqrt{1-x^2}} \\ = \dfrac{1}{1-x^2} * \dfrac{1}{\sqrt{1-x^2}} \\ \boxed{= \dfrac{\sqrt{1-x^2}}{x^4-2x^2+1}}$