SL 5.3—Differentiating polynomials, n E Z

The function $f(x)=5x^3-4x^2+7x-3$ has four terms: $5x^3$, $-4x^2$, $7x$, and $-3$.

For $5x^3$: multiply coefficient by power, then reduce power by 1. $5\times 3=15$, power becomes $2$, giving $15x^2$.

For $-4x^2$: $-4\times 2=-8$, power becomes $1$, giving $-8x$. For $7x$: the derivative of a linear term is just the coefficient, giving $7$.

The derivative of any constant is $0$, so $-3\to 0$.

$f^{\prime }(x)=15x^2-8x+7$

We need the derivative of a cubic polynomial. The correct answer must apply the power rule to every term, including the constant.

This option is missing the $+7$ term. The derivative of the linear term $7x$ is $7$, not $0$. This answer incorrectly drops the linear term.

The coefficient of $x$ is wrong. For $-4x^2$, the power rule gives $-4\times 2=-8$, so the term should be $-8x$, not $-4x$. This option forgot to multiply by the exponent.

The first term is wrong. For $5x^3$, the power rule gives $5\times 3=15$, not $5$. This option forgot to multiply the coefficient by the power.

$15x^2-8x+7$ correctly applies the power rule to every term: $5x^3\to 15x^2$, $-4x^2\to -8x$, $7x\to 7$, and $-3\to 0$.

We must differentiate $s(t)=4t^3-9t^2+6t+2$ to find $s^{\prime }(t)$, then substitute $t=2$.

Using the power rule term by term: $4t^3\to 12t^2$, $-9t^2\to -18t$, $6t\to 6$, $2\to 0$.

$s^{\prime }(t)=12t^2-18t+6$

$s^{\prime }(2)=12(2{)}^2-18(2)+6=12(4)-36+6=48-36+6=18$

The velocity at $t=2$ is $18$.

We need $s^{\prime }(2)$. First differentiate, then substitute. The correct derivative is $s^{\prime }(t)=12t^2-18t+6$.

This could come from forgetting to subtract correctly. $12(4)-18(2)+6=48-36+6=18$, not $30$. Likely an arithmetic error.

This may come from only computing $12(4)-36=12$, i.e. forgetting to add the $+6$ term from differentiating $6t$.

This could come from only evaluating the constant term of $s^{\prime }(t)$ or using a completely incorrect derivative.

$s^{\prime }(2)=12(4)-18(2)+6=48-36+6=18$, confirming the answer is $18$.