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

Answer 1

William Abdalla

#d/dx tan^-1(x-sqrt(1+x^2)) = (1-(x/sqrt(1+x^2)))/(1+(x-sqrt(1+x^2))^2#

#d/dx tan^-1(x) = 1/(1+x^2)#

Now, treat #x-sqrt(1+x^2)# as #x# in the above definition.

That would give us, #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)# is #1-(x/(sqrt(1+x^2)))#

Multiplying the derivative would give us,

#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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Answer 2

Adam Adkins

To differentiate ( y = \arctan(x - \sqrt{1 + x^2}) ), use the chain rule. The derivative is:

[ \frac{dy}{dx} = \frac{1}{1 + (x - \sqrt{1 + x^2})^2} \cdot (1 - \frac{x}{\sqrt{1 + x^2}}) ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.