# Find the derivative of y with respect to the appropriate variable?? y=tan^-1•√(x^2-1)+csc^-1•x, x>1 Thank You!

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### 1 Answer

You need to differentiate the function with respect to x, using chain rule, such that:

`(dy)/(dx) = 1/(1 + (sqrt(x^2-1))^2)*(sqrt(x^2-1))'*(x^2-1)' - 1/(|x|sqrt(x^2-1))`

`(dy)/(dx) = 1/(1 + x^2 - 1)*(1/(2sqrt(x^2-1)))*(2x) - 1/(|x|sqrt(x^2-1))`

`(dy)/(dx) = 1/x^2*(x/(sqrt(x^2-1))) - 1/(|x|sqrt(x^2-1))`

`(dy)/(dx) = 1/x*(1/(sqrt(x^2-1))) - 1/(|x|sqrt(x^2-1))`

If `x>0 => |x| = x` , hence, substituting x for |x| yields:

`(dy)/(dx) = 1/(x*sqrt(x^2-1)) - 1/(x*sqrt(x^2-1))`

`(dy)/(dx) = 0`

If `x<0 => |x| = -x` , hence, substituting -x for |x| yields:

`(dy)/(dx) = 1/(x*sqrt(x^2-1))+ 1/(x*sqrt(x^2-1))`

`(dy)/(dx) = 2/(x*sqrt(x^2-1))`

Hence, evaluating derivative of the given function, under the given conditions, yields `(dy)/(dx) = 0` (if x>0) or `(dy)/(dx) = 2/(x*sqrt(x^2-1))` (if x<0).