Question

How do you Find the derivative of `y=arctan(x-sqrt(1+x^2))`?

Answer 1

By Chain Rule,
`y'=1/{2(1+x^2)}`

Let us look at some details.

Let us first find the derivative of `x-sqrt{1+x^2}`.
By rewriting the square-root as the 1/2 power,
`(x-sqrt{1+x^2})'=[x-(1+x^2)^{1/2}]'`

by Chain Rule,
`=1-1/2(1+x^2)^{-1/2}cdot(2x)=1-x/sqrt{1+x^2}`

by taking the common denominator,
`={sqrt{1+x^2}-x}/sqrt{1+x^2}`

Now, we can find `y'`.
By Chain Rule,
`y'=1/{1+(x-sqrt{1+x^2})^2}cdot{sqrt{1+x^2}-x}/sqrt{1+x^2}`

by multiplying out the denominator of the first quotient,
`=1/{2(1+x^2-xsqrt{1+x^2})}cdot{sqrt{1+x^2}-x}/sqrt{1+x^2}`

by multiply the quotients together,
`={sqrt{1+x^2}-x}/{2[(1+x^2)sqrt{1+x^2}-x(1+x^2)]}`

by factoring out `(1+x^2)` from the denominator,
`={sqrt{1+x^2}-x}/{2(1+x^2)(sqrt{1+x^2}-x)}`

by cancelling `sqrt{1+x^2}-x`,
`=1/{2(1+x^2)}`