What is the instantaneous rate of change of f(x)=ln(4x^2-6) at x=-2 ?
1 Answer
Apr 5, 2016
Explanation:
This is the value of f'(-2)
differentiate using the
color(blue)" chain rule "
d/dx [f(g(x)) ] = f'(g(x)) . g'(x)
"-----------------------------------------------------------------" f(g(x))
= ln(4x^2 - 6) rArr f'(g(x)) = 1/(4x^2 - 6) and g(x)
=4x^2 - 6 rArr g'(x) = 8x
"----------------------------------------------------------------"
rArr f'(x) = 1/(4x^2 - 6) xx 8x = (8x)/(4x^2 - 6)
rArr f'(-2) = (-16)/(16-6) = -8/5
