SL 5.3—Introduction to derivatives

The function $f(x)=5x^4$ is a single power term, so apply the power rule $f^{\prime }(x)=an{x}^{n-1}$.

Here $a=5$ and $n=4$, so the new coefficient is $5\times 4=20$.

Subtract 1 from the exponent: $4-1=3$.

Combining these gives $f^{\prime }(x)=20x^3$.

$f^{\prime }(x)=5x^3$ forgets to multiply the coefficient by the exponent, only reducing the power.

$f^{\prime }(x)=20x^4$ multiplies correctly but forgets to reduce the exponent by 1.

$f^{\prime }(x)=9x^3$ mistakenly adds the coefficient and exponent ($5+4=9$) instead of multiplying.

Only $20x^3$ correctly applies both steps of the power rule.

$f(x)=-12$ is a constant function, which can be written as $-12x^0$.

Using $f^{\prime }(x)=an{x}^{n-1}$ with $n=0$ gives $-12\times 0\times x^{-1}=0$.

The exponent factor of 0 makes the entire expression zero regardless of $x$.

Therefore $f^{\prime }(x)=0$, consistent with the rule that constants always vanish upon differentiation.

$f^{\prime }(x)=-12$ mistakes the derivative for the original function value.

$f^{\prime }(x)=1$ has no basis in the power rule and does not follow from any correct step.

$f^{\prime }(x)=-12x$ incorrectly treats the constant as if it were a linear term $-12x$.

Only $f^{\prime }(x)=0$ correctly reflects that constants have zero rate of change.