What is the derivative of #f(x) = sqrt[ (3 x + 1) / (5 x^2 + 1) ]#?

1 Answer
Mar 27, 2018

Answer:

Use chain rule: if #f(x)=h(g(x))# the derivative is#f'(x)=h'(g(x))*g'(x)#
and quotient rule:
#(f/g)^'={\frac {f'g-fg'}{g^{2}}}# where #g!=0#

Explanation:

#f'(x)=(((3 x + 1)/ (5 x^2 + 1))^(1/2))^'#
#=1/2*((3 x + 1)/ (5 x^2 + 1))^(-1/2)*((3 x + 1)/ (5 x^2 + 1))^'#

Lets calculate #((3 x + 1)/ (5 x^2 + 1))^'# separately

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

#=(3 * (5 x^2 + 1)-(3 x + 1)*10x)/((5 x^2 + 1)^2)#
#=(15x^2+3-30x^2-10x)/((5 x^2 + 1)^2)=#
#=(-15x^2-10x+3)/((5 x^2 + 1)^2)#

The result is:
#f'(x)=(-15x^2-10x+3)/(2(5 x^2 + 1)^2 ((3 x + 1)/ (5 x^2 + 1))^(1/2)#