Racholg2Flbuze
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  • 01-10-2016
  • Mathematics
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What is the derivative of y=tan(arcsin(x))?

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caylus
caylus caylus
  • 01-10-2016
Hello,

(tg(x))'=1/cos²(x)

(arcsin(x))'=1/√(1-x²)

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


[tex](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}}\\ [/tex]
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