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IgorLugansk [536]
3 years ago
15

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

Mathematics
1 answer:
Anuta_ua [19.1K]3 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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We know that the area of a parallelogram is the base * the height of it, so if we divide both sides by the height, then area/height=base. Therefore, we must divide the area by the height. To divide using polynomials, we first set it up similar to a regular long division problem:


        ______________________

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Next, we take the first component of the numerator (2x² in this case) and divide it by the first component of the denominator (2x) to get x. That will form the start of our answer, and at the bottom, we will subtract our numerator by the denominator (2x+3) multiplied by the start of our answer (x). Therefore, we have


          _x_____________________

2x+3 | 2x²+13x+15

        -(2x²+3x)

        _______________

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_x+5_____________________

2x+3 | 2x²+13x+15

        -(2x²+3x)

        _______________

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0


Therefore, our base has a length of x+5.


Feel free to ask further questions, and Happy Holidays!

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