Question

question: find (dy) and evaluate when (x = 3) and (dx=0.4) for the function (y=-2x^{3}-1). provide your answer below: (dy=square)

Explanation:

Step1: Differentiate the function

The derivative of $y = - 2x^{3}-1$ with respect to $x$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$.
$y^\prime=\frac{dy}{dx}=-6x^{2}$

Step2: Calculate $dy$

We know that $dy = y^\prime dx$. Substituting $y^\prime=-6x^{2}$ and $dx = 0.4$, we get $dy=-6x^{2}dx$.

Step3: Evaluate when $x = 3$ and $dx = 0.4$

Substitute $x = 3$ and $dx = 0.4$ into the $dy$ formula.
$dy=-6\times3^{2}\times0.4$
$dy=-6\times9\times0.4$
$dy=-21.6$

Answer:

$dy=-21.6$