How do you find f'(x) and f''(x) given #f(x)=x^2/(1+2x)#?

Question

8624 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-01T03:51:54

#f'(x) = (2x(x+1))/((2x+1)^2)#

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

We're asked to find the first and second derivative of

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

First Derivative:

Use the quotient rule, which states

#d/(dx)[u/v] = (v(du)/(dx) - u(dv)/(dx))/(v^2)#

where

#= ((2x+1)d/(dx)[x^2] - x^2d/(dx)[2x+1])/((2x+1)^2)#

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

#= color(red)((2x^2+2x)/((2x+1)^2)#

Second derivative:

Use quotient rule again, this time

#= ((2x+1)^2d/(dx)[2x^2+2x] - (2x^2+2x)d/(dx)[(2x+1)^2])/((2x+1)^4)#

Use the chain rule on the #(2x+1)^2# term:

#= ((2x+1)^2(4x+2) - (2x^2+2x)(8x+4))/((2x+1)^4)#

Simplifying yields

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