Question

How do you differentiate `y=(x^2-2sqrtx)/x`?

Answer 1

`y'=(xsqrtx+1)/(xsqrtx)`

Explanation:

The function is differentiated by using the Quotient Rule differentiation

`y=(u(x))/(v(x))`

where `u(x)=x^2-2sqrtx and v(x)=x`

`color(red)(y'=(u'(x)v(x)-v'(x)u(x))/(v(x))^2)`

Computing `color(red)(y')`is determined by computing
`color(red)(u'(x)) and color(red)(v'(x))`

`u(x)` is a function of polynomials ,it is differentiated using the power rule differentiation rule:
`color(blue)((x^n)'=nx^(n-1))`

Computing `color(red)(u'(x))`
Knowing that :sqrtx=x^(1/2)#

`u(x)=x^2-2sqrtx`
`u'(x)=x^2-2x^(1/2)`
`u'(x)=color(blue)(2x^1)-2(color(blue)(1/2x^(-1/2)))`
`u'(x)=2x-1/(x^(1/2))`
`color(red)(u'(x)=2x-1/sqrtx)`

Computing `color(red)(v'(x))`
`v(x)=x`
`color(red)(v'(x)=1)`

`color(red)(y'=(u'(x)v(x)-v'(x)u(x))/(v(x))^2)`

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

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

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

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

`y'=((x^2sqrtx+x)/sqrtx)/x^2`

`y'=(x^2sqrtx+x)/(x^2sqrtx)`

`y'=(xsqrtx+1)/(xsqrtx)`

Answer 2

`dy/dx=1+1/(xsqrt(x))`

Explanation:

As an addition to the other answer, we can also simplify prior to differentiation, and then use the power rule.

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

`= x - 2/sqrt(x)`

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

`=> dy/dx = d/dx(x^color(red)(1)-2x^(color(red)(-1/2)))`

`=color(red)(1)x^(1-1) - 2(color(red)(-1/2))x^(-1/2-1)`

`=1x^0-(-1)x^(-3/2)`

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

`=1+1/(xsqrt(x))`

We could also put it back over a common denominator, giving us the same answer as we would have obtained from the quotient rule.

`=(xsqrt(x))/(xsqrt(x)) + 1/(xsqrt(x))`

`=(xsqrt(x)+1)/(xsqrt(x))`