What is the instantaneous rate of change of #f(x)=1/(x^2+2x+3 )# at #x=0 #?
1 Answer
Mar 10, 2016
Explanation:
This is just the value of f'(0)
# f(x) = 1/(x^2+2x+3) = (x^2+2x+3)^-1 # using the
#color(blue) " chain rule "#
#d/dx[f(g(x))] = f'(g(x)). g'(x)#
#rArr f'(x) = - (x^2+2x+3)^-2 d/dx(x^2+2x+3)#
# = -(2x+2)/(x^2+2x+3)^2 # and f'(0) =
#- 2/(3^2) = -2/9 #
