# Find dy/dx: `y= x(x^2+1)^-(1/2)``` Could someone find an answer for this differential problem and then show the work?

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### 1 Answer

`y= x(x^2+1)^-(1/2)`

`==> y= x/(x^2+1)^(1/2)`

Using differentiation rules, we know that:

If `f(x)= u/v ==> f'(x)= (u'v-uv')/v^2`

`==> y' = ((x)'(x^2+1)^(1/2) - (x)((x^2+1)^(1/2))')/((x^2+1)^(1/2))^2`

`==> y'= ((x^2+1)^(1/2) - (x)(1/2)(2x)(x^2+1)^-(1/2))/(x^2+1)`

`==> y'= ((x^2+1)^(1/2) - (x^2)(x^2+1)^(-1/2))/(x^2+1)`

`==> y'= (((x^2+1)-x^2)/(x^2+1)^(1/2))/(x^2+1)`

`==> y'= 1/((x^2+1)^(1/2) (x^2+1))= 1/(x^2+1)^(3/2)`

`==> y'= 1/(x^2+1)^(3/2)`

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