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How to do second derivative implicit differentiation using Wolfram Alpha?

Posted 5 years ago
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How do I perform second derivative implicit differentiation using Wolfram Alpha? I'm looking for results that will give me the step-by-step solutions as I'm a first year calculus student trying to figure things out.

I've tried all of the obvious queries that I can think of without getting the desired results. If I type just the equation in it will give the results I'm seeking but without the step-by-step solution since it is not the primary output for the query.

Thanks for any assistance.

Here's an example of the results I get from just entering the equation x^2 + xy = 5 using Wolfram Alpha input in Mathematica (w/desired result circled):

enter image description here

P.S. Cross posted to http://mathematica.stackexchange.com/questions/51705/second-derivative-implicit-differentiation-using-wolfram-alpha-input

5 Replies

Hi Gary,

this probably does not really answer your question. I am also not quite sure what your input was and I cannot reproduce what you get, but if you do the calculation in two steps it works. First you take the first derivative

enter image description here

then calculate another derivative of the first result.

enter image description here

As you see you get to see step-by-step solutions. The second solution contains the first and second derivative. You then need to substitute the first result:

y'[x] + x y''[x] == y[x]/x  /. {y'[x] -> -2 - y[x]/x}

The only thing that remains is to solve for y''[x].

Solve[%, y''[x]]

From you screenshot I see that you have Mathematica and not only Wolfram Alpha, so this recipe should work. You have to do a bit of work by hand, but you get to see all the other intermediate steps shown.

I am not sure how to get this with step-by-step solutions with one query to Wolfram Alpha. What was your input?

Cheers, M.

Posted 5 years ago

Whoops. FYI - I replied at the end of the page instead of hitting the reply button on your post.

Posted 5 years ago
Posted 5 years ago

Hi Marco,

Thanks for the response.

The step-by-step on the second equation is helpful until about halfway through when it starts doing some weird algebra that ends up with y'' mixed in with the other terms. Another weird thing is that the end result of the step-by-step is different than the initial result shown on the main page (i.e. it solves for y').

I'm a newbie to Mathematica and have been trying to mix in learning its syntax along with my other work. I'm curious about the following bit; can you break this down for me just a little bit -- mainly what the "/." does?

y'[x] + x y''[x] == y[x]/x  /. {y'[x] -> -2 - y[x]/x}

I tried a number of things to get this to work so I can't remember them all. I do recall using the following 2 queries:

(d^2 y(x))/(dx^2) x^2+ xy(x)=5

second derivative x^2+xy(x)=5

I'm surprised that there isn't an easily discovered way to do this since it obviously can calculate y'' as evidenced by the results I got from just entering the equation by itself. I wish that there was more documentation on the recognized syntax but I imagine that based on the wide-ranging nature of the Wolfram Alpha engine that it would require volumes of it so that's probably too much to ask.

Thanks again,
Gary

Dear Gary,

there is a lot of notation in Mathematica which might be unfamiliar at first but is very powerful. Most of the symbols are really useful. If you type

? /. 

so a question mark in front of the symbol you don't know you get a lot of info on it. If you then click on the >> symbol you go to the help page. You can also do this by highlighting the symbol and then clicking, e.g. Fn+F1 on a Mac, and something similar on Windows.

/. 

means "replace all". The help page shows how to use it and lots of examples. The examples are interactive so you can try out what the command does. These pages are really useful.

In the example above (last post)

y'[x] + x y''[x] == y[x]/x  /. {y'[x] -> -2 - y[x]/x}

you can read the /. symbol as "such that". So use the equation such that the replacement rules are taken into consideration. So it basically just substitutes y'[x] by -2-y[x]/x.

It is quite important to learn the (few and logical) rules behind the Wolfram Language. You will see in this community that you can do an awful lot of cool stuff with it. The linguistic input and Wolfram Alpha are very useful, but it is also important to learn the syntax behind the language. There are excellent tutorials on the Wolfram Website as well:

http://www.wolfram.com/training/courses/

Lot's of them are on demand and free! I understand though that the step-by-step solution is a very nice feature and that you want to use that for as many problems as possible.

Hope this helps, Marco

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