How do you find the derivative of #f(x) = [(3x^2 + 1)^(1/3) - 5]^2 / [5x^2 + 4]^(1/2)#?

1 Answer
Aug 24, 2016

#f'(x)=[(4x(5x^2+4)^(1/2))/((3x^2+1)^(2/3)){(3x^2+1)^(1/3)-5}-(5x)/(5x^2+4)^(1/2){(3x^2+1)^(1/3)-5}^2]/(5x^2+4)#

Explanation:

We will use the Quotient Rule and Chain Rule.

#f(x)={(3x^2+1)^(1/3) -5}^2/(5x^2+4)^(1/2)=u/v, say,#, where,

#u={(3x^2+1)^(1/3)-5}^2, &, v=(5x^2+4)^(1/2)#.

Now, #f=u/v rArr f'=(vu'-uv')/v^2...............(star)#.

#u={(3x^2+1)^(1/3)-5}^2#

#rArr u'=2{(3x^2+1)^(1/3)-5}{(3x^2+1)^(1/3)-5}'#

#=2{(3x^2+1)^(1/3)-5}{(3x^2+1)^(1/3)}'#

#=2{(3x^2+1)^(1/3)-5}{1/3(3x^2+1)^(-2/3)}(3x^2+1)'#

#=2/3*6x{(3x^2+1)^(1/3)-5}(3x^2+1)^(-2/3)#

#u'=(4x)/(3x^2+1)^(2/3){(3x^2+1)^(1/3)-5}#

#v=(5x^2+4)^(1/2)#.

#rArr v'=1/2(5x^2+4)^(-1/2)(5x^2+4)'#

#=1/2*10x(5x^2+4)^(-1/2)#

#v'=(5x)/(5x^2+4)^(1/2)#

Using all these, together with, #v^2=(5x^2+4)#, in #(star)#, we get,

#f'(x)=[(4x(5x^2+4)^(1/2))/((3x^2+1)^(2/3)){(3x^2+1)^(1/3)-5}-(5x)/(5x^2+4)^(1/2){(3x^2+1)^(1/3)-5}^2]/(5x^2+4)#

Enjoy maths.!