What is the instantaneous rate of change of f(x)=ln(4x^2+2x ) at x=-1 ?

1 Answer
Ken C.
Jun 1, 2016

-3

Explanation:

Instantaneous rate of change is simply the derivative. To find it, take the derivative of the function and evaluate it at the desired x-value.

We have a logarithmic function with a polynomial inside, which means we need to use the chain rule. As it applies the the natural log function, the chain rule is:
d/dx(ln(u))=(u')/u
Where u is a function of x.

In this case, u=4x^2+2x, so u'=8x+2. Therefore,
f'(x)=(8x+2)/(4x^2+2x)=(2(4x+1))/(2(2x^2+x))=(4x+1)/(2x^2+x)

All that's left to find instantaneous rate of change is to evaluate this at x=-1:
f'(-1)=(4(-1)+1)/(2(-1)^2+(-1))=-3/1=-3

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