Declaring variables for use on Calculus deriving program

[Matt Morin]

I'm pretty new to programming and I'm trying to create a deriving program to do derivatives. For now, I would only like to know how I could declare a variable such as derivative to be the equation where the user inputs what each variable is. I am currently just trying to develop a product rule calculating program. I am having trouble do to getting the actual letter x into the equation so I could display the answer as "45x^(3)+12" or such. Is it possible to do so the way I'm going? Or is there a different way to approach this problem? Or maybe would I have to tell the user to input a value for x and the output would be a plain number?

[Matt Morin]

If I was to require the input of an x for power rule, why does the equation a*n*(x^(n-1)) with variables a=2, n=3, x=1 come out to be 18? Here's my Power Rule Code:

#include <iostream>
using namespace std;
int a, x, n, derivative;
int main ()
{
cout<< "Hello. I am a computer program. I can calculate the derivative of an expression using power rule.";
cout<< "\nPlease type the values of a and n for use like the ex.:";
cout<< "\ny=ax^n";
cin.get ();
cout<< "a=?\n";
cin>> a;
cout<< "x=?\n";
cin>> x;
cout<< "n=?\n";
cin>> n;
derivative=a*n*(x^(n-1));

cout<< "dy/dx=" << derivative;
return 0;
}

[ResidentBiscuit]

^ Is the bit wise XOR operator, not the power operator. You'll have to use the math library for a power function.

[Matt Morin]

I found this out on my own by comparing the binary codes and what the answers I was getting. Thank you, though! I just didn't see your post in time. I found that you have to include math.h and example command is pow(2, 3) which would come out to 8. Would you happen to know anything about my original question?

[ResidentBiscuit]

I'm not really sure what you're asking up there. Are you wanting to display the actual 'x' in the equation, or the value of x in the equation? Either way it pretty simple

[Matt Morin]

I want to display x so that x's power could be shown and answers such as 24x+32 or whatever could be shown. Can it (not sure what to call it, the computer? processor?) process an equation with a letter inside and give an answer such as the example, 24x+32? I will post my revised power rule of having to input a value for x.

[Matt Morin]

//Power Rule Program
#include <iostream>
#include <math.h>
using namespace std;

int a, x, n, y, derivative;

int main ()
{
cout<< "Hello. I am a computer program. I can calculate the derivative of an expression using power rule.";
cout<< "\nPlease type the values of a and n for use like the ex.:";
cout<< "\ny=ax^n";
cin.get ();
cout<< "a=?\n";
cin>> a;
cout<< "x=?\n";
cin>> x;
cout<< "n=?\n";
cin>> n;

y=pow(x, (n-1));
derivative=a*n*y;

cout<< "dy/dx=" << derivative;
return 0;
}

[Matt Cromer]

First of all, always enter your code using the code tool:

//Power Rule Program
#include <iostream>
#include <math.h>
using namespace std;

int a, x, n, y, derivative;

int main ()
{
cout<< "Hello. I am a computer program. I can calculate the derivative of an expression using power rule.";
cout<< "\nPlease type the values of a and n for use like the ex.:";
cout<< "\ny=ax^n";
cin.get ();
cout<< "a=?\n";
cin>> a;
cout<< "x=?\n";
cin>> x;
cout<< "n=?\n";
cin>> n;

y=pow(x, (n-1));
derivative=a*n*y;

cout<< "dy/dx=" << derivative;
return 0;
}

Second of all, i think this is over my head.
Haven't even learned calculus yet... XD

[Matt Morin]

Didn't know there was a code tool, thanks. Well, it would be helpful if you could help me figure out how to get something simple like:

//Equation simplifier
#include <iostream>
using namespace std;

int a, b, c, d, x, answerleft, answerright; //I know x couldn't be an integer in this case, correct?

int main ()
{
    cout<< "I can simplify equatinos of format 'ax+c=bx+d'.\nPlease enter the values for a, b, c, and d.\n";
    cin.get ();
    cout<< "a=?\n";
    cin>> a;
    cout<< "c=?\n";
    cin>> c;
    cout<< "b=?\n";
    cin>> b;
    cout<< "d=?\n";
    cin>> d;

    answerleft=a*x-b*x;
    answerright=d-c;

    cout<< answerleft << "=" <<answerright;
    return 0;
}

I would like to be able to plug-in (ex.) a=3, c=2, b=7, and d=12 and have a return answer of -4x=10 not 0=10.

[ResidentBiscuit]

Well part of your problem is, x is never initialized. Not sure why your using it in that equation. I believe anymore, compilers default non-initialized numerics to 0, so your left side will always be 0. If it doesn't initialize for you, it'll just be garbage. Either way, it's bad.

[Bibek Subedi]

If you simply display answerleft, it will display Integer value but not alphabetic numerals. So you cant display x because it is not integer, it is actually a variable name. I think you want to convert x (but not its value) to String or char type and display it as a variable in an equation. But i think it is very difficult in C/C++ (but i am not sure).

[Matt Cromer]

Hmmm perhaps if you know that x is always in the same place you could work around it..

cout << answerleftofx << "x" << answerrightofx << "=" << answerright;

Something like that maybe?

[Matt Cromer]

//Equation simplifier
#include <iostream>
using namespace std;

int a, b, c, d, answerleft, answerright; //I know x couldn't be an integer in this case, correct?

int main ()
{
    cout<< "I can simplify equatinos of format 'ax+c=bx+d'.\nPlease enter the values for a, b, c, and d." << endl;
    cout<< "a= ";
    cin>> a;
    cout<< "c= ";
    cin>> c;
    cout<< "b= ";
    cin>> b;
    cout<< "d= ";
    cin>> d;

    answerleft=a-b;
    answerright=d-c;

    cout<< answerleft << "x=" <<answerright;
    return 0;
}

[Matt Morin]

I know what my problem is in the code and I can work around it' but there are equations that don't work as well when trying to work around. Bibek Subedi is correct in what I was asking and trying to do.