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")

% from Stewart's Calc 6 ed p. 599 just examples 1 and 2
%INPUT

syms f c t y(t) y yprime yIntegral;
y=(1+c*exp(t))/(1-c*exp(t))

y = 

simplify(y)

ans = 

yprime=diff(y)

yprime = 

%f=int(f);
yprime=simplify(yprime)

yprime = 

%disp(f);
yIntegral=int(yprime);
simplify(yIntegral);
disp(yIntegral);

OUTPUT

-(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]

You are not specifying the variable of differentiation, and you are not specifying the variable of integration.

[John D'Errico]

@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"

Answer 1

[Paul]

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

symvar(y,1)

ans = 

t

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

symvar(yprime,1)

ans = 

t

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 = 

[Joseph Palumbo]

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!