# Differentiate y = (sqrt x + x)/x^2

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### 2 Answers

The derivative of `y = (sqrt x + x)/x^2` has to be determined.

Now for the function `y = (sqrt x + x)/x^2` divide each term of the numerator by the denominator x^2.

This gives:

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

= `x^(-3/2) + x^-1`

The derivative of `y = x^n` is `y' = n*x^(n-1)`

For `y = x^(-3/2) + x^-1`

`y' = (-3/2)*x^(-5/2) - 1*x^(-2)`

**The derivative of `y = (sqrt x + x)/x^2` is **`y'= (-3/2)*x^(-5/2) - 1*x^(-2)`

### User Comments

To differentiate the function `y = (sqrt x + x)/x^2` use the quotient rule of differentiation.

For `y = (f(x))/(g(x))` , the quotient rule gives `y' = (f'(x)*g(x) - f(x)*g(x))/(g(x))^2` .

The derivative of `y = (sqrt x + x)/x^2` is:

`y' = ((sqrt x + x)'x^2 - (sqrt x + x)*(x^2)')/(x^2)^2`

= `((1/(2*sqrt x) + 1)*x^2 - (sqrt x + x)*(2x))/(x^4)`

= `(x^2/(2*sqrt x) + x^2 - 2*x*sqrt x - 2x^2)/(x^4)`

= `(x^2/(2*sqrt x) - x^2 - 2*x*sqrt x)/(x^4)`

= `(x/(2*sqrt x) - x - 2*sqrt x)/(x^3)`

= `(x - x*2*sqrt x - 4 x)/(2*sqrt x*x^3)`

= `(-3x - x*2*sqrt x)/(2*sqrt x*x^3)`

= `(-3 - 2*sqrt x)/(2*sqrt x*x^2)`

The required derivative is `(-3 - 2*sqrt x)/(2*sqrt x*x^2)`