# Finding a differential

I am trying to compute $\Delta y$ and dy for a given value of x and dx = $\Delta x$ I am given $y=2x-x^2$ x=2 and $\Delta x = -.04$

Not really sure what to do here, feels like something is missing. I thought I knew how to do these problems but they took another approach to it than what I am used to.

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Do your notes indicate $\Delta y=f(x+\Delta x)-f(x)$ and $dy=f'(x)dx$? – anon Oct 6 '11 at 0:49
No, Stewart doesn't make that distinction very clear. I am attempt to decipher and transcribe his writings right now. – user138246 Oct 6 '11 at 1:16

$\Delta y$ means the change in $y$ when $x$ changes by $\Delta x$, so it's the value of $y$ at $x+\Delta x$ minus the value of $y$ at $x$. Since you know the formula for $y$, and you are given the values of $x$ and of $\Delta x$, you should be able to get $\Delta y$.

If we write $f'(x)$ for the derivative, then I'm assuming $dy$ is being defined by $dy=f'(x)dx$, so again all you have to do is compute the derivative at the given value of $x$ and then plug in.

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I can't wrap my head around this stuff at all. I know what it looks like on a graph but I can't think of why that works in a function. Maybe math isn't for me. – user138246 Oct 6 '11 at 1:18

On the graph of $y=f(x)$, the finite difference $\Delta y$ can be found thusly: look at the two points $(x,f(x))$ and $(x+\Delta x,f(x+\Delta x))$ on the graph, note $\Delta y$ is the change in $y$ component from the first to the second, so we get $\Delta y=f(x+\Delta x)-f(x)$.

As for the differential $dy$, you can look at the derivative $dy/dx=f'(x)$ and formally multiply by $dx$ to get $dy=f'(x)dx$. Does this make sense to you?

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You've been given a function y=f(x) and told to find $\Delta y$ (the change in $y$). Specifically the change in $y$ as $x$ changes from $x=2$ to $x=2+\Delta x=2-0.4=1.6$ . So the change in $y$ is $\Delta y = f(1.6)-f(2)$.

On the other hand $dy$ is an approximation of the change in $y$ using the tangent line. Keep in mind that the tangent line at $x=2$ is given by $y-f(2) = f'(2) (x-2)$. This says the change in $y$ is approximately $f'(2)$ times the change in $x$ (this is approximate because we're using a tangent and not the function itself). Thus $\Delta y \approx dy = f'(2)dx$.

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