Derivative of y = 3 ^ (4x + 1) The process I forgot to do it several times

Derivative of y = 3 ^ (4x + 1) The process I forgot to do it several times

(a^x)'=a^x*lna
So y '= 3 ^ (4x + 1) * Ln3 * (4x + 1)'
=4*3^(4x+1)*ln3

Given that y = FX is an odd function defined on R, when x is greater than 0, FX = the square of X - 3x + 2, then FX=

A:
F (x) is an odd function on R: F (- x) = - f (x), f (0) = 0
x>0,f(x)=x^2-3x+2
x<0,-x>0：f(-x)=(-x)^2-3(-x)+2=x^2+3x+2=-f(x)
So: X & lt; 0, f (x) = - x ^ 2-3x-2
So:
            x^2-3x+2,x>0
  f(x)= { 0,            x=0
            -x^2-3x-2,x<0

Find the n-order derivative of function y = (4x ^ 2 + x) / (1-x-2x ^ 2)

y=（4x^2+x）/(1-x-2x^2)=（4x^2+2x-2-x+2）/(1-x-2x^2)=-2+（x-2）/(x+1)(2x-1)=-2+1/(x+1)-1/(2x-1)∴y'=-1/(x+1)^2+2/(2x-1)^2y''=2/(x+1)^3-2*4/(2x-1)^2……………… Y (n order) = (- 1) ^ n * n! / (x + 1) ^ (n + 1) - (- 2) ^ n * n! / (2