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

Question

23237 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-11T23:28:03

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)}#