DylsexicChciken Posted March 24, 2014 Share Posted March 24, 2014 (edited) For the second derivative, why is (d/dx)(dy/dx)= [(d/dt)(dy/dx)]/(dx/dt)? I don't see the logic. Am I supposed to apply the quotient rule of differentiation on dy/dx? If I do that I get the second derivative is 0 because dy-dy=0. Am I allowed to use algebra on dy and dx? Edited March 24, 2014 by DylsexicChciken Link to comment Share on other sites More sharing options...

John Posted March 24, 2014 Share Posted March 24, 2014 (edited) What's the context in which this equation comes about? Using the Leibniz notation, I guess the way to think about it is [math]\frac{d}{dx} \frac{dy}{dx} = \frac{d \frac{dy}{dx}}{dx} = \frac{d \frac{dy}{dx}}{dx} \frac{\frac{1}{dt}}{\frac{1}{dt}} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}}[/math]. In general we can just consider "multiplying" the differentials, obtaining [math]\frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} \frac{d}{dx} y = \frac{d^{2}}{dx^{2}} y = \frac{d^{2} y}{dx^{2}}[/math], which is the standard Leibniz notation for the second derivative of [math]y[/math] with respect to [math]x[/math]. Leibniz notation is handy for seeing the process of taking derivatives, in certain cases, and in some limited sense we can, as above, talk about "multiplying" and "dividing" and so forth. However, it's not really valid (at least, at the introductory standard calculus level) to treat differentials as quantities that can be manipulated algebraically willy-nilly. Edited March 24, 2014 by John Link to comment Share on other sites More sharing options...

ajb Posted March 24, 2014 Share Posted March 24, 2014 You understand x now as a function of t and y is a function of x and apply the chain rule. Loosely you can treat dx, dy and dt as indeterminate and use standard rules of algebra, in particular they commute and you can divide by them. But I will stress, this is only a formal handling of objects and many mathematicians don't like it. Link to comment Share on other sites More sharing options...

DylsexicChciken Posted March 24, 2014 Author Share Posted March 24, 2014 (edited) What's the context in which this equation comes about? Using the Leibniz notation, I guess the way to think about it is [math]\frac{d}{dx} \frac{dy}{dx} = \frac{d \frac{dy}{dx}}{dx} = \frac{d \frac{dy}{dx}}{dx} \frac{\frac{1}{dt}}{\frac{1}{dt}} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}}[/math]. In general we can just consider "multiplying" the differentials, obtaining [math]\frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} \frac{d}{dx} y = \frac{d^{2}}{dx^{2}} y = \frac{d^{2} y}{dx^{2}}[/math], which is the standard Leibniz notation for the second derivative of [math]y[/math] with respect to [math]x[/math]. Leibniz notation is handy for seeing the process of taking derivatives, in certain cases, and in some limited sense we can, as above, talk about "multiplying" and "dividing" and so forth. However, it's not really valid (at least, at the introductory standard calculus level) to treat differentials as quantities that can be manipulated algebraically willy-nilly. Ah, so the author of my book multiplied the top and bottom by (1/dt)/(1/dt). The book didn't say you can "multiply" and the book just did this, which is really uninformative: [math]\frac{d}{dx} \frac{dy}{dx} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}}[/math] Which skipped a lot of steps that would have helped me understand the notation better. Thanks for the help. ...But I will stress, this is only a formal handling of objects and many mathematicians don't like it. Yea, I can see why. I already dislike this notation. Edited March 24, 2014 by DylsexicChciken Link to comment Share on other sites More sharing options...

John Posted March 24, 2014 Share Posted March 24, 2014 Ah, so the author of my book multiplied the top and bottom by (1/dt)/(1/dt). The book didn't say you can "multiply" and the book just did this, which is really uninformative: [math]\frac{d}{dx} \frac{dy}{dx} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}}[/math] Which skipped a lot of steps that would have helped me understand the notation better. Thanks for the help. Well, I should say I don't know for sure that the author did it that way, heh, or indeed what his thought process was at all. That's why I was wondering what the context was. This was simply the process that occurred to me to get from the left-hand side to the right-hand side. Link to comment Share on other sites More sharing options...

ajb Posted March 24, 2014 Share Posted March 24, 2014 Ah, so the author of my book multiplied the top and bottom by (1/dt)/(1/dt). He multiplied by 1, in effect. Yea, I can see why. I already dislike this notation. You are not alone in that. The notation and formal handling can be useful, but it gets confusing when you really try to understand what is going on, which is best described in terms of limits. Also later on this notation gets confused with other objects usually also denoted as dx. Link to comment Share on other sites More sharing options...

studiot Posted March 24, 2014 Share Posted March 24, 2014 I think that the difficulty arises at least in part becuse you are thinking that manipulation in John's post#2 and in your book implies multiplication. It does not. It implies composition of functions. The output of composition is another function. I am going to assume you understand what a function is, please ask if you are unsure as this is an important concept. In your notation y is a function of x. We can obtain a second function properly called "the derived function" by following certain rules, known as differentiation. The derived function is often abbreviated to the derivative. However you should realise it is a function in its own right. Moreover it is a function of x (not y). So we can attempt to obtain a derived function from it. I say attempt because not all derived functions are suitable to provide a derived function. If successful, the second derived function will also be a function of x. A pity you are not being taught by my old A level maths teacher. He was the best teacher I have ever had in any subject. He used to insist that when we performed differentiation we made clear the variable concerned by writing "differentiate with respect to x" or d.w.r.x for short. You always got a mark for that even if you fluffed the algebra of the differentiation itself. This practice is good in more advanced work since it helps keep track of things. To recap, differentiation operates on functions and its output is a function. Consequenctly we can repeat this operation multiple times, following the rules for composition of functions each time. Link to comment Share on other sites More sharing options...

DylsexicChciken Posted March 26, 2014 Author Share Posted March 26, 2014 (edited) Oh yea, I forgot to mention something that may be important. In [latex] \frac{d}{dx} \frac{dy}{dx} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}} [/latex] The variable t is a parameter for both y and x. So x and y can are functions of t. Meanwhile, y is a function of x. I forgot to mention this as I thought it would not be important. My main trouble is with understanding how to "operate" in Leibniz notation since I have never learned any rules to manipulate the Leibniz notation. E.g., you learn properties of commutation and distribution and etc, which you can apply to arithmetic, but I have no idea how to start operating in Leibniz notation. For example, how to even begin expanding: [latex] \frac{d}{dx} \frac{dy}{dx}[/latex] Given that x and y can are functions of t, and y is a function of x. I would have never known the above can be expanded to the second derivative: [latex]\frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}} [/latex] If no one told me so. I understand how to get the first derivative, you have by the chain rule of differentiation: [latex] \frac{dy}{dt}=\frac{dy}{dx}*\frac{dx}{dt}[/latex] The book divides both sides of the equation by [latex]\frac{dx}{dt}[/latex] to get: [latex] \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} [/latex] But how would you start calculating the second derivative: [latex]\frac{d}{dx}\bigg(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\bigg) [/latex] I have tried applying the quotient rule of differentiation, which I assume is the wrong approach since I keep getting 0 for the numerator, what do you do to the above to get: [latex]\frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}} [/latex] Edited March 26, 2014 by DylsexicChciken Link to comment Share on other sites More sharing options...

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