AHL 5.9—Differentiating standard functions and derivative rules

Rewrite $f(x)=\frac{5}{x^3}$ as $f(x)=5x^{-3}$ so the power rule can be applied directly.

Using $\frac{d}{dx}(x^n)=n{x}^{n-1}$ with $n=-3$: $f^{\prime }(x)=5(-3)x^{-4}$.

$f^{\prime }(x)=-15x^{-4}=-\frac{15}{x^4}$.

The derivative is $-\frac{15}{x^4}$.

$-\frac{15}{x^2}$ comes from reducing the power by only 1 unit incorrectly on the exponent's sign, mixing up $-4$ with $-2$.

$\frac{15}{x^4}$ has the correct magnitude but drops the negative sign from multiplying by $n=-3$.

$-\frac{5}{x^2}$ forgets to multiply the coefficient 5 by the exponent $-3$, and misapplies the new power.

Only $-\frac{15}{x^4}$ correctly applies both the coefficient multiplication and power reduction.

$\sqrt[3]{x}=x^{1/3}$.

Using $n=\frac{1}{3}$: derivative is $\frac{1}{3}{x}^{1/3-1}$.

$\frac{1}{3}-1=-\frac{2}{3}$, giving $\frac{1}{3}{x}^{-2/3}$.

The derivative is $\frac{1}{3}{x}^{-2/3}$.

$3x^{2/3}$ inverts the coefficient (using $\frac{1}{n}$ instead of $n$) and keeps the wrong sign on the exponent.

$\frac{1}{3}{x}^{2/3}$ keeps the correct coefficient but fails to subtract 1 from the exponent correctly, giving a positive exponent.

$x^{-2/3}$ has the correct exponent but omits the coefficient $\frac{1}{3}$ entirely.

Only $\frac{1}{3}{x}^{-2/3}$ correctly reduces the exponent by 1 and retains the coefficient.