If $m(x)=\frac{g(3x-2)}{4x}$, what is the instantaneous rate of change at $x=2$?

Key concept: Rewrite the function as a product of $g(3x-2)$ and $x^{-1}$, then use the chain rule and product rule together.

Step by step

Step 1: Rewrite the function

$m(x)=\frac{g(3x-2)}{4x}=\frac14\,g(3x-2)\,x^{-1}$

Step 2: Differentiate

Use the product rule on $g(3x-2)\,x^{-1}$:

$m'(x)=\frac14\left[(g(3x-2))'x^{-1}+g(3x-2)(x^{-1})'\right]$

Now apply the chain rule:

$(g(3x-2))'=3g'(3x-2)$

and

$(x^{-1})'=-x^{-2}$

So,

$m'(x)=\frac14\left[3g'(3x-2)x^{-1}-g(3x-2)x^{-2}\right]$

Step 3: Evaluate at $x=2$

$m'(2)=\frac14\left[3g'(4)\cdot\frac12-g(4)\cdot\frac14\right]$

$m'(2)=\frac{3}{8}g'(4)-\frac{1}{16}g(4)$

Final answer

$\boxed{m'(2)=\frac{3}{8}g'(4)-\frac{1}{16}g(4)}$

Pitfall alert

Do not forget that both the numerator and denominator contribute to the derivative. A common error is treating $4x$ as a constant or skipping the chain rule on $g(3x-2)$.

Try different conditions

If the denominator were $4x^2$ instead of $4x$, then the power of $x$ would change and the second term in the derivative would be different. The chain-rule part, however, would still use $g'(3x-2)$ multiplied by 3.

Further reading

quotient rule, chain rule, instantaneous rate of change