What is the instantaneous rate of change of f(x)=ln⁡(4x2+2x) at x=-1?

Question

Jeremiah Dickerson · Accepted Answer

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:

\frac{d}{d x} (\ln (u)) = \frac{u^{'}}{u}

Where u is a function of x.
In this case,

u = 4 x^{2} + 2 x

, so u'=8x+2. Therefore,

f^{'} (x) = \frac{8 x + 2}{4 x^{2} + 2 x} = \frac{2 (4 x + 1)}{2 (2 x^{2} + x)} = \frac{4 x + 1}{2 x^{2} + x}

All that's left to find instantaneous rate of change is to evaluate this at x=-1:

f^{'} (- 1) = \frac{4 (- 1) + 1}{2 {(- 1)}^{2} + (- 1)} = - \frac{3}{1} = - 3

Answer:
-3

What is the instantaneous rate of change of

Answered question

Answer & Explanation

New Questions in Calculus 1