How do you differentiate #h(x)=((x^3+1)sqrtx)/x^2# using the quotient rule?

2 Answers
Alan N.
Mar 16, 2018

#h'(x) = (3sqrtx(x^3-1))/(2x^3)#

Explanation:

#h(x) = ((x^3+1)sqrtx)/x^2#

#= (x^3+1)/x^(3/2)#

Apply the quotient rule.

#h'(x) = (x^(3/2)* (3x^2) - (x^3 +1) * (3/2x^(1/2)))/x^3#

#= (3x^(7/2) - 3/2sqrtx(x^3+1))/x^3#

#= (6x^(7/2) - 3sqrtx(x^3+1))/(2x^3)#

#= (6x^(7/2)-3x^(7/2) -3sqrtx)/(2x^3)#

#= (3x^(7/2) - 3sqrtx)/(2x^3)#

#= (3sqrtx(x^3-1))/(2x^3)#

maganbhai P.
Mar 16, 2018

#h^'(x)=(3(x^3-1))/((2sqrtx)x^2)=3/2*x^(1/2)-3/2*x^(-5/2)#, (simplify)
OR
#h(x)=((x^3+1)sqrt(x))/x^2=(x^3sqrtx)/x^2+(sqrtx)/x^2=x^(3+1/2-2)+x^(1/2-2)=x^(3/2)+x^(-3/2)=>h^'(x)=3/2*x^(1/2)-3/2*x^(-5/2)#

Explanation:

Here, #h(x)=((x^3+1)sqrt(x))/x^2#
Let,#f(x)=(x^3+1)sqrtxandg(x)=x^2=>g^'(x)=2x#
#andf^'(x)=(x^3+1)d/(dx)(sqrtx)+sqrtx*d/(dx)((x^3+1)#
#=>f^'(x)=(x^3+1)*1/(2sqrtx)+sqrtx*3x^2=(x^3+1+2x*3x^2)/(2sqrtx)#
#=>f^'(x)=(x^3+1+6x^3)/(2sqrtx)=(7x^3+1)/(2sqrtx)#
We know that, If #h(x)=(f(x))/(g(x)),then,#
#h^'(x)=(g(x)*f^'(x)-f(x)*g^'(x))/([g(x)]^2)=(x^2((7x^3+1)/(2sqrtx))-(x^3+1)sqrt(x)*2x)/((x^2)^2)=(x^2(7x^3+1)-(x^3+1)2x*2x)/((2sqrtx)*x^4)#
#=(7x^5+x^2-(x^3+1)4x^2)/((2sqrtx)x^4)=(7x^5+x^2-4x^5-4x^2)/((2sqrtx)x^4)=(3x^5-3x^2)/((2sqrtx)x^4)=(3x^2(x^3-1))/((2sqrtx)x^4)=(3(x^3-1))/((2sqrtx)x^2)#

Related questions
See all questions in Quotient Rule
Impact of this question
2172 views around the world
You can reuse this answer
Creative Commons License