What is the instantaneous rate of change of #f(x)=1/(x^3-2x+5 )# at #x=0 #?
1 Answer
Apr 15, 2016
Explanation:
This is just the value of the derivative at x = 0.
differentiate using the
#color(blue)" chain rule "#
#d/dx [ f(g(x)) ] = f'(g(x)) . g'(x) # Rewrite
# 1/(x^3 - 2x + 5) = (x^3 - 2x + 5)^-1 #
#"------------------------------------------------------"# f(g(x)) =
#(x^3-2x+5)^-1 #
#rArr f'(g(x)) = -1(x^3-2x+5)^-2 # and g(x) =
#x^3-2x+5 rArr g'(x) = 3x^2-2 #
#"----------------------------------------------------------"#
now substitute these values into the derivative
#rArr f'(x) = -(x^3-2x+5)^-2 xx (3x^2-2)#
# = (-(3x^2-2))/(x^3-2x+5)^2 # and f'(0)
# =( -(-2))/(5)^2 = 2/25 #
