How do you find the derivative of $y = arctan (\sqrt{3 (x^{2}) - 1})$ $y = arctan(sqrt(3(x^2)-1))$ ?

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How do you find the derivative of $y = arctan (\sqrt{3 (x^{2}) - 1})$ $y = arctan(sqrt(3(x^2)-1))$ ?

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Answer 1 · 2017-06-18T10:11:30

$\frac{d y}{d x} = \frac{1}{x \sqrt{3 x^{2} - 1}}$ $dy/dx=1/(xsqrt(3x^2-1))$

Explanation:

The first thing to know is the derivative of the function $y = {tan}^{- 1} (u)$ $y=tan^-1(u)$ , where $u = g (x)$ $u=g(x)$ (think "the function on the inside").

$\frac{d}{d x} ({tan}^{- 1} (u)) = \frac{1}{1 + u^{2}} \frac{d u}{d x}$ $d/dx(tan^-1(u))=1/(1+u^2) (du)/dx$

$\frac{d y}{d x} = \frac{d}{d x} [{tan}^{- 1} (\sqrt{3 x^{2} - 1})]$ $dy/dx=d/dx[tan^-1(sqrt(3x^2-1))]$

$= \frac{1}{1 + {(\sqrt{3 x^{2} - 1})}^{2}} \frac{d}{d x} [{(3 x^{2} - 1)}^{\frac{1}{2}}]$ $=1/(1+(sqrt(3x^2-1))^2) d/dx[(3x^2-1)^(1/2)]$

$= \frac{1}{1 + 3 x^{2} - 1} (\frac{1}{2} {(3 x^{2} - 1)}^{- \frac{1}{2}} (6 x))$ $=1/(1+3x^2-1) (1/2(3x^2-1)^(-1/2)(6x))$

$=1/(cancel(3)x^cancel(2)) (cancel(3)cancel(x)(3x^2-1)^(-1/2))$

$=1/(xsqrt(3x^2-1))$

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How do you find the derivative of $y = arctan (\sqrt{3 (x^{2}) - 1})$ $y = arctan(sqrt(3(x^2)-1))$ ?

1 Answer

Explanation:

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How do you find the derivative of y = arctan(sqrt(3(x^2)-1))y=arctan(√3(x2)−1)y = arctan(sqrt(3(x^2)-1))?

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Explanation:

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How do you find the derivative of $y = arctan (\sqrt{3 (x^{2}) - 1})$ $y = arctan(sqrt(3(x^2)-1))$ ?