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

Question

2162 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-27T18:24:54

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

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 · 2016-10-27T18:46:26

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

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))#