How do you differentiate $y = arctan (x - \sqrt{1 + x^{2}})$ $y= arctan(x - sqrt(1+x^2))$ ?

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How do you differentiate $y = arctan (x - \sqrt{1 + x^{2}})$ $y= arctan(x - sqrt(1+x^2))$ ?

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Answer 1 · 2016-06-20T22:39:06

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

Explanation:

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

Now, treat $x - \sqrt{1 + x^{2}}$ $x-sqrt(1+x^2)$ as $x$ $x$ in the above definition.

That would give us,
$\frac{d}{d x} {tan}^{- 1} (x - \sqrt{1 + x^{2}}) = \frac{1}{1 + {(x - \sqrt{1 + x^{2}})}^{2}}$ $d/dx tan^-1(x-sqrt(1+x^2)) = 1/(1+(x-sqrt(1+x^2))^2$

Don't forget the chain rule though!

The derivative of $x - \sqrt{1 + x^{2}}$ $x-sqrt(1+x^2)$ is $1 - (\frac{x}{\sqrt{1 + x^{2}}})$ $1-(x/(sqrt(1+x^2)))$

Multiplying the derivative would give us,

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

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How do you differentiate $y = arctan (x - \sqrt{1 + x^{2}})$ $y= arctan(x - sqrt(1+x^2))$ ?

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Explanation:

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How do you differentiate y=arctan(x−√1+x2)y= arctan(x - sqrt(1+x^2))?

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How do you differentiate $y = arctan (x - \sqrt{1 + x^{2}})$ $y= arctan(x - sqrt(1+x^2))$ ?