What is the instantaneous rate of change of #f(x)=ln(4x^2-6) # at #x=-2 #?

1 Answer
Jim G.
Apr 5, 2016

# -8/5 #

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 #

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