Question

Let y=e^x. Find the differential dy when x = 4 and dx = 0.2.

dy = ___.

Find the differential dy when x = 4 and dx = 0.02.

dy = ___.

Answer 1

\[\frac{ dy }{dx }=e ^{x},dy=e ^{x}dx\]
plug the values of x and dx

Answer 2

Sry didn't see you posted

Answer 3

Thanks! It worked!

Answer 4

It would be (e)^(4)*(0.2), and (e)^(4)*(0.02).

Answer 5

\[0.02 e ^{4}\]

Answer 6

@surjithayer
Well... Honestly, I gotta say something about this is really bothering me.
Is that really the correct solution???
Maybe I don't get the question right, or having a mistake..
Let me explain what's bothering me (sorry if it's long):

Wikipedia: http://en.wikipedia.org/wiki/Differential_(infinitesimal)
"The term differential is used in calculus to refer to an infinitesimal (infinitely small) change in some varying quantity."

So differential like \( dx \) is defined as infinitely small change on the \( x \).
Since in our function the \( y \) depends on the \( x \) then change on the \( x \) implies a change on the \( y \).
If the function is differentiable at that point then this change on the \( y \) would also be infinitely small.
And then, the ratio between the infinitely small differentials \( \Large \frac{dy}{dx} \) gives the slope of the curve on that point.
Therefore, if \( f(x) \) is differentiable at \( x \) we can say:
\[
\lim_{\Delta x \to 0}{ \frac{\Delta y}{\Delta x} } = \frac{dy}{dx} = f'(x) \\
\frac{dy}{dx} = f'(x) \quad \implies \quad dy = f'(x) \cdot dx
\]
The problem is.. that we can't say \(dx = 0.2\) as \(dx\) is a differential.. which is defined as infintely small. So let's use \( \Delta x = 0.2 \) instead.

But \( \Large \frac{ \Delta y }{ \Delta x } \) isn't necessarily \( f'(x) \) as \( \Delta x \) is not infintely small..
So if you're trying to find \( \Delta y \) (change on the \( y \)) for \( \Delta x \) (the change on the \( x \)) then simply:
\[
\Delta y = f(x + \Delta x) - f(x)\\
\Delta y = f(4 + 0.2) - f(4) \\
\Delta y = e^{4.2} - e^{4} = 66.69 - 54.60 = 12.09
\]
While \(\Delta y = 0.2 \cdot e^4= 0.2 \cdot 54.60 = 10.92 \)
Am I missing anything?

Answer 7

dy is only an approximation of \[\Delta x\]

Answer 8

Dy and dx aren't always infinitely small, either...

Answer 9

But if \(\Delta x\) (and therefore \(\Delta y\)) is not infinitely small then you can't replace \( \frac{\Delta y}{\Delta x} \) with the derivative.. Am I right?

Answer 10

Well, dy and dx are only approximations, see? Near the tangent line, they closely approximate \[\Delta x\] and \[\Delta y\].

Answer 11

I don't want to bother too much, but I'll try one more time to understand...

When I say \(dx\) I mean the differential that as I know is defined to be infinitesimal (like wikipedia says).
While \(\Delta x\) just represents a change on the x, and if it's infinitely small then it becomes \(dx\).
Thing is, I know this definition to derivative:
\[
f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx}
\]
And if \(\Delta x \) is not infinitely small as in the diffinition, then it doesn't give the derivative..

Let me give an example... think of \(f(x) = x^2 \)
The derivative is of course \(f'(x) = 2x \)
Let's take point \(x=0\) and say \(\Delta x = 2\)
If we say
\[
\frac{\Delta y}{\Delta x} = f'(x) \quad \implies \quad \Delta y = f'(x) \Delta x
\]
Then we get
\[
\Delta y = f'(0) \cdot 2 = (2 \cdot 0) \cdot 2 = 0
\]
But if we just check...
\[
f(x) = f(0) = 0^2 = 0 \\
f(x + \Delta x) = f(0 + 2) = (0+2)^2 = 4 \\
\Delta y = f(x + \Delta) - f(x) = 4 - 0 = 4
\]
The change of the function's value along \(\Delta x\) from \(x\) in this case is 4...