user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rdi
  lea    D60(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x38,%rsp
  call   <puts@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
  lea    D63(%rip),%rdi
  call   <puts@plt>
  movsd  D5(%rip),%xmm2
  movsd  D94(%rip),%xmm5
  nopw   0x0(%rax,%rax,1)
L4:
  movapd %xmm5,%xmm0
  movsd  %xmm2,0x18(%rsp)
  movsd  %xmm5,0x8(%rsp)
  call   <f@plt>
  movsd  0x8(%rsp),%xmm5
  movsd  %xmm0,0x10(%rsp)
  movapd %xmm5,%xmm0
  call   <diff@plt>
  movsd  0x10(%rsp),%xmm1
  movsd  0x8(%rsp),%xmm5
  movsd  0x18(%rsp),%xmm2
  divsd  %xmm0,%xmm1
  movapd %xmm5,%xmm4
  movsd  %xmm2,0x10(%rsp)
  movsd  %xmm4,0x20(%rsp)
  subsd  %xmm1,%xmm5
  movapd %xmm5,%xmm2
  movsd  %xmm5,0x28(%rsp)
  subsd  %xmm4,%xmm2
  andpd  D19(%rip),%xmm2
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x18(%rsp)
  call   <log@plt>
  movsd  0x10(%rsp),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm1,%xmm0
  call   <log@plt>
  movsd  0x8(%rsp),%xmm3
  movsd  0x20(%rsp),%xmm4
  divsd  %xmm0,%xmm3
  movapd %xmm4,%xmm0
  movsd  %xmm4,0x8(%rsp)
  movsd  %xmm3,0x10(%rsp)
  call   <f@plt>
  movsd  0x8(%rsp),%xmm4
  mov    %ebx,%esi
  mov    %rbp,%rdi
  movsd  0x18(%rsp),%xmm2
  movsd  0x10(%rsp),%xmm3
  movapd %xmm0,%xmm1
  add    $0x1,%ebx
  movapd %xmm4,%xmm0
  mov    $0x4,%eax
  movsd  %xmm2,0x10(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm4
  movsd  0x28(%rsp),%xmm5
  movsd  D62(%rip),%xmm6
  movsd  0x10(%rsp),%xmm2
  subsd  %xmm5,%xmm4
  andpd  D19(%rip),%xmm4
  comisd %xmm4,%xmm6
  jbe    L4
  movapd %xmm5,%xmm0
  lea    D77(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
type_infered: string
name: unpack_data
arguments: tag: D60
type_infered: string
name: unpack_data
arguments: tag: D1
type_infered: string
name: unpack_data
arguments: tag: D63
type_infered: string
name: unpack_data
arguments: tag: D5
type_infered: f64
name: unpack_data
arguments: tag: D94
type_infered: f64
name: unpack_data
arguments: tag: D19
type_infered: qword[2]
name: unpack_data
arguments: tag: D62
type_infered: f64
name: unpack_data
arguments: tag: D77
type_infered: string

tool: D3: Solution for e^x = 2x+1 using newton raphson method

tool: D60: %d	%f	%f	%f	%f


tool: D1: Initial approximation:1

tool: D63: i	Root		f(Root)		Error		Order of convergence

tool: D5: 2

tool: D94: 1

tool: {"D19": [0x7fffffffffffffff,0x0000000000000000]}

tool: D62: 1e-06

tool: D77: 

Final Answer:%f


assistant: ```c
int main() {
  printf("Solution for e^x = 2x+1 using newton raphson method\n");
  printf("Initial approximation:1\n");
  double newRoot, oldRoot = 1;
  double oldError = 2, newError;
  int i = 0;
  printf("i\tRoot\t\tf(Root)\t\tError\t\tOrder of convergence\n");
  while (1) {
    newRoot = oldRoot - (f(oldRoot) / diff(oldRoot));
    newError = fabs(newRoot - oldRoot);
    printf("%d\t%f\t%f\t%f\t%f\n", i++, oldRoot, f(oldRoot), newError, log(newError) / log(oldError));
    if (fabs(oldRoot - newRoot) < 0.000001)
      break;
    oldRoot = newRoot;
    oldError = newError;
  }
  printf("\n\nFinal Answer:%f\n", newRoot);
  return 0;
}
```