Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 22

Calculate $\Delta y$ and $dy$ of $y = x^3$ for $x = 1$ and $ \Delta x = 0.5$. Then sketch a diagram showing the line segments with lengths $dx$, $dy$, and $\Delta y$.

Solving for $\Delta y$


$
\begin{equation}
\begin{aligned}
f(x + \Delta x) f(1 + 0.5) = f(1.5) &= (1.5)^3\\
\\
f(1.5) &= 3.375
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
f(x) = f(1) &= (1)^3\\
f(1) &= 1
\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}
\Delta y &= f(1.5) - f(1)\\
\Delta y &= 3.375 -1\\
\Delta y &= 2.375
\end{aligned}
\end{equation}
$


Solving for $dy$

$
\begin{equation}
\begin{aligned}
dy &= f'(x) dx\\
\\
\frac{dy}{dx} &= \frac{d}{dx} (x^3)\\
\\
dy &= 3x^2 dx\\
\\
dy &= 3(1)^2(0.5)\\
\\
dy &= 1.5
\end{aligned}
\end{equation}
$