Question

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

Answer 1

`-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`