First off, no, this is not homework. This comes from self-study and has stymied me. Please explain your answer as thoroughly as you can!
Find increment $\Delta y$ and differential $dy$ for the function $y=x^3$ if $x=1$ and $\Delta x = 1$
If you need any more detail please ask.
On
The increment $\Delta y$ is the change in the function when $x$ changes by $\Delta x$. Thinking of $y$ as $f(x)$, we want $\Delta y = f(x+\Delta x)-f(x)$. That formula is general, for any function. In this case, our function is $f(x)=x^3$, so the formula becomes:
$$\begin{align}\Delta y &= (x+\Delta x)^3 - x^3 \\ &= x^3+3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3 - x^3 \\ &= 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3\end{align}$$
We're interested in the case where $x=1$, so we substitute that in and obtain $\Delta y=3\Delta x + 3(\Delta x)^2+(\Delta x)^3$.
Also, we have that $\Delta x=1$, so we finally obtain $\Delta y=3+3+1=7$.
When thinking about the differential, we let $\Delta x$ become vanishingly small, which makes $\Delta y$ vanishingly small as well. Higher order terms disappear entirely, so we end up with $dy=3x^2dx$, or when $x=1$, this becomes $dy=3dx$. Taking $dx=1$ gives us $dy=3$.
Does that answer your question?
Well, by implicit differentiation, it is pretty clear that since $y=x^3$, $dy=3x^2dx$. So, at $x=1$, $\boxed{dy=3dx}$.
$\Delta y$ is the change in $y$. So, let $\Delta x$ be the change in $x$. So, $$\Delta y=(x+\Delta x)^3-x^3\\ \Delta y=3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3$$ At $x=1$, $$\boxed{\Delta y=3\Delta x+3(\Delta x)^2+(\Delta x)^3}$$ As $\Delta x=1$, $$\boxed{\Delta y=3+3+1=7}$$ Also, $\lim_{\Delta x\to 0}dy=3\lim_{\Delta x\to0}dx\approx3\Delta x=3$. So, $\boxed{dy\approx 3}$.