How do you find the first and second derivatives of #(2x)/(sqrt(5x^2 -2x + 1))# using the quotient rule?

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Answer 1 · 2015-10-27T15:36:50

The first derivative is #f'(x)=(2-2x)/(5x^2-2x+1)^(3/2)# and the second derivative is #f''(x)=(20x^2-32x+4)/(5x^2-2x+1)^(5/2)#.

The quotient rule says #d/dx(g(x)/(h(x)))=(h(x) * g'(x) - g(x) * h'(x))/((h(x))^2)#.

For #f(x)=(2x)/sqrt(5x^2-2x+1)#, we get

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

#=(10x^2-4x+2-(10x^2-2x))/(5x^2-2x+1)^(3/2)=(2-2x)/(5x^2-2x+1)^(3/2)#

and then

#f''(x)=((5x^2-2x+1)^(3/2) * (-2)-(2-2x) * 3/2 (5x^2-2x+1)^(1/2) * (10x-2))/((5x^2-2x+1)^3)#

#=((5x^2-2x+1)^(1/2) * (-10x^2+4x-2-3(1-x)(10x-2)))/((5x^2-2x+1)^3)#

#=(-10x^2+4x-2+30x^2-36x+6)/((5x^2-2x+1)^(5/2))#

#=(20x^2-32x+4)/(5x^2-2x+1)^(5/2)#.