Double derivative by c++

[mehat]

//How to find the value of 2nd derivative.
#include <iostream>
#include <cmath>
using namespace std;
int main(){
int po,p,mul;
int x1,x2,x3,diff=0,x,n=1;
cout<<("enter the value of x at which you want to calculate derivative = ")<<endl;
cin>>x;
for(;;)
{
cout<<("enter power of ") <<n<< (" term = ")<<endl;
cin>>p;
cout<<("enter multiplier of ")<<n<<(" term = ")<<endl;
cin>>mul;
n=n+1;
if(p==0 )
{
break;
}
else
{
po=p-2;
x1=p*mul;
x2=pow(x,po);
x3=x2*x1;
diff=diff+x3;}}
cout<<"the diffrential of polinomial is = "<<diff;
return 0;}

[TomCPP]

The 2nd derivative of polynomial axⁿ is n(n-1)x^n-2 ( for n<2 is zero ).
You should apply this calculation formula to each parts of the following equation:
a₁xⁿ + a₂x^n-1 + ... + a_n-1x + a_n

C++ speaking, you should collect both degree of polynomial and coefficients a_n,..,a₁. Then calculate the 2nd derivative form of polynomial which reduces its degree by two. Finally calculate the value.

[lastchance]

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

using Poly = vector<double>;

Poly deriv( Poly p )
{
   if ( p.size() <= 1 ) return Poly{ 0 };
   for ( int i = 1; i < p.size(); i++ ) p[i-1] = i * p[i];
   p.resize( p.size() - 1 );
   return p;
}

double evaluate( const Poly &p, double x )
{
   double result = 0.0;
   for ( int i = p.size() - 1; i >= 0; i-- ) result = p[i] + x * result;
   return result;
}

void write( const Poly &p )
{
   cout << p[0];
   for ( int i = 1; i < p.size(); i++ ) if ( p[i] ) cout << ( p[i] > 0 ? " + " : " - " ) << abs( p[i] ) << "x^" << i;
}

int main()
{
   Poly y = { 2, -4, 1, 3 };    // 2 - 4x + x^2 + 3x^3
   Poly Dy = deriv( y );
   Poly D2y = deriv( Dy );

   double x = 3;
   cout << "x = " << x << '\n';
   cout << "Original poly:     ";   write( y   );   cout << " = " << evaluate( y  , x ) << '\n';
   cout << "First derivative:  ";   write( Dy  );   cout << " = " << evaluate( Dy , x ) << '\n';
   cout << "Second derivative: ";   write( D2y );   cout << " = " << evaluate( D2y, x ) << '\n';
}

x = 3
Original poly:     2 - 4x^1 + 1x^2 + 3x^3 = 80
First derivative:  -4 + 2x^1 + 9x^2 = 83
Second derivative: 2 + 18x^1 = 56

[againtry]

struct's come in handy too:

#include <iostream>

struct Term // ax^n
{
    int a{0};
    int n{0};
    
    Term n_th_derivative(int level = 1)
    {
        Term result{a,n};
        int N = n;
        
        for(int i = 0; i < level; i++)
        {
            result.a *= N;
            
            N--;
            result.n = N;
        }
        return result;
    }
    
    std::string display()
    {
        if(a != 0)
            return std::to_string(a) + " x^ " + std::to_string(n);
        else
            return "Zero";
    }
};

int main()
{
    int a{0}, n{0};
    
    std::cout
    << "A single term has the form aX^n\n"
    << "Enter a: ";
    std::cin >> a;
    
    std::cout
    << "Enter n: ";
    std::cin >> n;
    
    int d{2};
    Term term{a,n};
    std::cout
    << "The second derivative of " << term.display() << " is "
    << term.n_th_derivative(d).display() << '\n';
    
    return 0;
}

A single term has the form aX^n
Enter a: 9
Enter n: 5
The second derivative of 9 x^ 5 is 180 x^ 3
Program ended with exit code: 0

[againtry]

And as an alternative to <vector>'s and subject to tidy-up terms can be arrayed to become equations:

#include <iostream>

struct Term // ax^n
{
    int a{0};
    int n{0};
    
    Term n_th_derivative(int level = 1)
    {
    Term result{a,n};
    int N = n;
    
    for(int i = 0; i < level; i++)
        {
        result.a *= N;
        
        N--;
        result.n = N;
        }
    return result;
    }
    
    std::string display()
    {
    if(a != 0 and n !=0)
        return std::to_string(a) + " x^ " + std::to_string(n);
    else
        return std::to_string(a);
    
    return "Zero";
    }
};

int main()
{
    Term equation[]{ Term{2,0}, Term{-4,1}, Term{1,2}, Term{3,3} };
    int no_terms = sizeof(equation)/sizeof(Term);
    
    for (int i = 0; i < no_terms - 1; i++)
    std::cout << equation[i].n_th_derivative(2).display() << " + ";
    
    std::cout << equation[no_terms - 1].n_th_derivative(2).display() << " = ";
    
    return 0;
}