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 2019-01-11, 02:50 #145 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 1BC16 Posts We're starting to do a bit on differentiation, with "a bit" being emphasized. I'm having trouble figuring out why $$x^3$$ is not differentiable at x = 0. If the derivative of $$x^3$$ is defined here, doesn't that mean that it is differentiable? Or do I have the definition wrong? I took a look at definite integrals, and it's neat that $\int_{0}^{2\pi} \sin(x)dx = 0.$I get how it works, but it's kinda fascinating! I wasn't sure how to do the whole integral notation thing in $$\LaTeX$$, so I used this nice online editor: https://www.latex4technics.com/ Other than an ad in the upper left, it's pretty cool. It translates code to output in real time, has preset buttons that insert code templates, and autocomplete if you can't remember commands. And you can export whatever is compiled as an image! Last fiddled with by jvang on 2019-01-11 at 02:51 Reason: typing is hard
 2019-01-11, 04:20 #146 VBCurtis     "Curtis" Feb 2005 Riverside, CA 507410 Posts Polynomials are differentiable everywhere, including x3 at x=0.
2019-01-11, 08:44   #147
Nick

Dec 2012
The Netherlands

3×587 Posts

Quote:
 Originally Posted by jvang If the derivative of $$x^3$$ is defined here, doesn't that mean that it is differentiable?
Yes, you're right. Perhaps you were thinking of $$\sqrt[3]{x}$$?
This is not differentiable at 0 as the tangent line to the graph at x=0 is vertical.

2019-01-11, 12:22   #148
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

11×17×59 Posts

Quote:
 Originally Posted by Nick Yes, you're right. Perhaps you were thinking of $$\sqrt[3]{x}$$? This is not differentiable at 0 as the tangent line to the graph at x=0 is vertical.
It's worse than that, it's dead Jim.

Not only is $$\sqrt[3]{x}$$ undefined at the origin, it changes sign there too, as does the first derivative as x approaches zero.

To see why its worse, think about trying to differentiate a step function where y = -1 if x<0, y=0 if x=0 and y=1 if x>0. This one doesn't run off to infinity anywhere and it is defined for all real x. It can be differentiated anywhere except at x=0 to give a function which is zero everywhere except at x=0.

Ok, you say, points of discontinuity are obvious reasons why a function can't be differentiated there. However, that's not the whole story. Consider now the function y=abs(x) defined as y=-x for x<0, y=x for x>=0. Its graph is a V-shaped curve and the function is continuous at the origin. However dy/dx =-1 for x<0 and +1 for x>0. Both limits exist as y->0 from above and below but they have different values. Accordingly dy/dx is undefined at the origin.

Your class teacher is unlikely to tell you about functions like these but, IMO, you should know about them because it helps you to understand that the value of dy/dx at x=a is the limit of [y(a+d) - y(a) / d] as d tends to zero from all directions. If y is undefined at x=a, or if the expression just given differs according to how the limit is taken, y is said to not to be differentiable at y=a.

You may wish to consult https://en.wikipedia.org/wiki/Differentiable_function for more detail but please don't be intimidated by the depth of treatment in that article. Most people meet it, if at all, in their first year of study for a mathematics degree.

2019-01-11, 18:14   #149
Dr Sardonicus

Feb 2017
Nowhere

5,147 Posts

Quote:
 Originally Posted by xilman It's worse than that, it's dead Jim. Not only is $$\sqrt[3]{x}$$ undefined at the origin, it changes sign there too, as does the first derivative as x approaches zero.
Yes, when you're working with real variables, non-integer powers generally can only be well defined for positive real numbers, using the real logarithm, and

$x^{r}\;=\;e^{(r\ln(x))}$.

With positive rational powers r with odd denominators, you can extend to all real numbers by defining 0^r as 0 and (-x)^r as - (x^r) for positive x. The function is then continuous everywhere and also differentiable, except, if r < 1, at 0. Negative powers obviously can not be defined at 0.

Unfortunately, the natural log goes completely haywire at 0. (Just how haywire, is way beyond the scope of an introductory calculus course.) What it means is, any attempt to extend noninteger powers beyond the positive real numbers using the above relation is fraught with peril.

Powers with rational exponents with even denominators can not be extended to real-valued functions of negative real numbers. Negative real numbers don't have real square roots.

At least with rational exponents r, the "ambiguity" in r-th powers is finite, being a factor of a root of unity. This can lead to fallacies even with the square root, where carelessness can lead to a "wrong square root" that is off by a factor of -1.

With irrational exponents, things are infinitely more complicated.

Last fiddled with by Dr Sardonicus on 2019-01-11 at 18:19

 2019-01-12, 18:19 #150 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22×3×37 Posts This Friday, we did some weird derivative stuff where we take something like: $x^2 + y^2 = 4$And take the derivative of it, resulting in: $2x+2y*y = 0$Then we solve for $$y$$ algebraically. Is there a special name for this, and how is it important? I can't see a use case for it, partially because I don't really know what it is in the first place
2019-01-12, 19:10   #151
Dr Sardonicus

Feb 2017
Nowhere

120338 Posts

Quote:
 Originally Posted by jvang This Friday, we did some weird derivative stuff where we take something like: $x^2 + y^2 = 4$And take the derivative of it, resulting in: $2x+2y*y = 0$Then we solve for $$y$$ algebraically. Is there a special name for this, and how is it important?
When I took calculus, it was called "implicit differentiation." It is valid at any point for which the expression defines y as a differentiable function of x in some neighborhood of the point. This is generally valid at any point where the expression for y' "makes sense."

It can be handy because you don't have to "solve" for y as a function of x. It also generally gives the derivative as a fraction (in your example, y' = -x/y). Note that the expression -x/y is undefined when y (the denominator) is 0, and the points on the circle where y = 0 are where the tangent line is vertical. (Note also that these points are where the two functions of x defining the top half and the bottom half of the circle meet.)

You might try implicit differentiation with fractional-powers functions defined implicitly, for example y = x^(2/3) can be expressed as y^3 = x^2 or y^3 - x^2 = 0.

 2019-01-17, 02:20 #152 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22×3×37 Posts I'm getting the hang of doing the implicit differentiation stuff, but the concept is a bit weird for me to understand. Our teacher has been absent for a couple days, so hopefully she'll explain this stuff more clearly tomorrow Here's a weird one I worked out involving trigonometric functions: $x+ \tan(xy) = 0$$1+\sec^2(xy)(xy'+y) = 0$$xy' = -\dfrac{1}{\sec^2(xy)}-y$$xy' = -\cos^2(xy) - y$$y' = -\dfrac{\cos^2(xy)-y}{x}$ It doesn't seem to difficult, except that I wouldn't know when to use it unless the question explicitly asked for it (our teacher does sometimes ask for specific solutions to certain test/exam problems)
 2019-01-17, 03:28 #153 VBCurtis     "Curtis" Feb 2005 Riverside, CA 507410 Posts You use it any time you can't reasonably isolate y in the original equation. If implicit differentiation is easier than isolating y, use it. There are many equations where isolating y is not possible; those need implicit method always. There are application exercises where the input variable is time, but the equation is a relationship among variables that are not time. Those equations always need implicit method too; they're often referred to as "related rates" problems. It pays to get comfortable with the notation and style of these problems; the method is used in 3 or 4 sections spread throughout the year's course.
 2019-01-18, 01:25 #154 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22·3·37 Posts There was a weird problem we worked with in class that no one was sure what to do with (teacher was as unsure as we were ). $\lim_{x\rightarrow \infty} \dfrac{\cos x}{x}$Logically speaking, the answer looks like it'll be 0. With x going to positive infinity, the denominator will be infinitely larger than what the numerator will be, which is somewhere between -1 and 1. But since we can't determine what exact number that $$\cos x$$ will be when x goes to infinity, how can we work with it for a mathematically rigorous proof? Our teacher mentioned the squeeze theorem, but the only way we've used it is for $$\dfrac{\sin x}{x}$$ and $$\dfrac{1-\cos x}{x}$$. Is it possible to generalize the squeeze theorem to this sort of end behavior problem?

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