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IgorLugansk [536]

2 years ago

15

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

1 answer:

Anuta_ua [19.1K]2 years ago

4 0

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}}\\

$

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Answer:

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Step-by-step explanation:

A first order function has the following format:

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Step-by-step explanation:

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