want to differentiate 1+c*exp(t)/1-exp(t) which is 2*c*exp(t)/(1-exp(t))^2 and then return it to it's original form by integration using int(), understand rewrite(tanh,"exp")

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% from Stewart's Calc 6 ed p. 599 just examples 1 and 2
syms f c t y(t) y yprime yIntegral;
y = 
ans = 
yprime = 
yprime = 
-(c*exp(t) + 1)/(c*exp(t) - 1)
-(c*exp(t) + 1)/(c*exp(t) - 1)
(c*exp(t)*(c*exp(t) + 1))/(c*exp(t) - 1)^2 - (c*exp(t))/(c*exp(t) - 1)
(2*c*exp(t))/(c*exp(t) - 1)^2
-2/(c*exp(t) - 1)
Walter Roberson
Walter Roberson on 23 Feb 2024
You are not specifying the variable of differentiation, and you are not specifying the variable of integration.
John D'Errico
John D'Errico on 26 Feb 2024
Edited: John D'Errico on 26 Feb 2024
@Joseph Palumbo - I moved your comment that you posted as the accepted answer. Since you accepted the answer, I could not even move it directly. Please learn to use comments.
"Please forgive me everyone, this is my favorite calculus text I studied in college however, it appears I misunderstood something, this part begins with y=(1+c*exp(t))/(1-c*exp(t)) including the c's which are constants only to show you this is a family of functions, each different 'c' completing a separate function of which they are all of the same family. These c's should be replaced by their respective constant before differentiating or integrating, what fooled me however is that MatLab did differentiate it properly even with the c's included: 2ce^t/(ce^t-1)^2, being the correct answer (matches the book) but when it comes to integrating it, c's definitely cannot be included they are only added in for the integration, itself to symbolize constants. I MUST REVALUATE MY QUESTION AND GET BACK!------Joseph Palumbo"

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Accepted Answer

Paul on 26 Feb 2024
Edited: Paul on 26 Feb 2024
Hi Joseph,
I think you got the correct result, though not using what might be considered best practices
syms f c t y(t) y yprime yIntegral
y = (1+c*exp(t))/(1-c*exp(t))
y = 
yprime = simplify(diff(y))
yprime = 
When calling diff without a second argument, the differentiation is taken wrt
ans = 
which in this case does happen to be the variable you want. To be safe, and for clarity, its probably better to always specify the variable of differentation, especially in an expression with more than one variable
yprime = simplify(diff(y,t))
yprime = 
Similarly, int w/o the second argument will integrate wrt to
ans = 
Again, it's the variable we want, but better to just be explicit
yr = int(yprime,t)
yr = 
which is the result you obtained.
However, int only returns an anti-derivative with a "constant of integration" being zero. So let's add a constant
syms K
yr = yr + K
yr = 
Now, we can solve for the value of K such that yr matches y
K = solve(yr == y,K)
K = 
Sub in that value of K to yr
yr = subs(yr)
yr = 
And simplify to the expected fraction
[num,den] = numden(subs(yr));
yr = num/den
yr = 
  1 Comment
Joseph Palumbo
Joseph Palumbo on 27 Feb 2024
Moved: Paul on 27 Feb 2024
Thanks alot Paul you explained alot of things that I am trying to do, which in the long run, is just to become familiar wth symbolic toolbox!

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