# Given that y = x-2/(x^2+5) , find dy/dx and determine the range of values of x for which both y and dy/dx are postitve .Please show me step by step and clarify how you did it through...

Given that y = x-2/(x^2+5) , find dy/dx and determine the range of values of x for which both y and dy/dx are postitve .

**Please show me step by step and clarify how you did it through differentiation**

**I didn't even get the quesition.**

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You need to differentiate with respect to x the given expression `y = (x-2)/(x^2+5)` , using the quotient rule such that:

`(dy)/(dx) = ((x-2)'(x^2+5) - (x-2)(x^2+5)')/((x^2+5)^2)`

You need to differentiate with respect to x the following terms:

`(x-2)' = (d(x-2))/(dx) = x' - 2' = 1 - 0 =1`

`(x^2+5)' = (d(x^2+5))/(dx) = (x^2)' + 5' = 2x + 0 =2x`

`(dy)/(dx) = (x^2 + 5 - 2x(x - 2))/((x^2+5)^2)`

`(dy)/(dx) = (x^2 + 5 - 2x^2 + 4x)/((x^2+5)^2)`

`(dy)/(dx) = (-x^2 + 4x + 5)/((x^2+5)^2)`

Notice that the denominator is positive for all values of x.

You should determine the range where `-x^2 + 4x + 5 > 0` such that:

`-x^2 + 4x + 5 = 0 => x^2 - 4x - 5 = 0`

Using quadratic formula `x_(1,2) = (-b+-sqrt(b^2-4ac))/(2a)` yields:

`x_(1,2) = (4+-sqrt(16 + 20))/2 => x_(1,2) = (4+-sqrt36)/2`

`x_(1,2) = (4+-6)/2 => x_1 = 5 ; x_2 = -1`

Hence, the expression -`x^2 + 4x + 5 > 0` if `x in (-1,5)` .

**Hence, evaluating `(dy)/(dx)` yields `(dy)/(dx) = (-x^2 + 4x + 5)/((x^2+5)^2)` and `(dy)/(dx) > 0` if `x in (-1,5).` **