How to find the slope of y = g(x) / x^3 at a specific point

Key concept: A quotient like $g(x)/x^3$ is easiest to handle by rewriting it or using the quotient rule. Either way, the derivative must be simplified before plugging in the point.

Step by step

We need the slope of

$y=\frac{g(x)}{x^3}$

at $x=2$.

Step 1: Rewrite if helpful

Write it as

$y=g(x)x^{-3}$

Step 2: Differentiate

Use the product rule:

$y' = g'(x)x^{-3} + g(x)(-3)x^{-4}$

So

$y' = \frac{g'(x)}{x^3} - \frac{3g(x)}{x^4}$

Step 3: Evaluate at $x=2$

$y'(2)=\frac{g'(2)}{2^3}-\frac{3g(2)}{2^4}$

which simplifies to

$y'(2)=\frac{g'(2)}{8}-\frac{3g(2)}{16}$

That is the slope of the tangent line at $x=2$.

Pitfall alert

Don’t differentiate only the numerator and leave the denominator alone. The $x^3$ in the denominator also contributes to the derivative. A second trap is substituting $x=2$ too early; simplify first, then evaluate.

Try different conditions

If the denominator were $x^n$, the same pattern would be

$\frac{d}{dx}\left(\frac{g(x)}{x^n}\right)=\frac{g'(x)}{x^n}-\frac{ng(x)}{x^{n+1}}$

at points where $x\neq 0$.

Further reading

quotient rule, slope of tangent line, product rule