Find the derivative of $\sin { \left( 3{ x }^{ 2 }+x \right) } $ How do you find the derivative of

$$\sin { \left( 3{ x }^{ 2 }+x \right)  } $$ 

using the derivative definition and not the chain rule.
This is how far I was able to get
Any help would be appreciated. Thanks!
 A: Select $u=3x^2+x\Rightarrow \frac{du}{dx}=6x+1$ then $$\frac{d}{dx}(\sin(u))=\frac{du}{dx}\cos(u)=(6x+1)\cos(3x^2+x)$$
A: \begin{eqnarray*}\lim _{ h\rightarrow0 }&{ \frac { \sin { \left( 3{ \left( x+h \right)  }^{ 2 }+x+h \right)  } -\sin { \left( 3{ x }^{ 2 }+x \right)  }  }{ h }  } \\=&\lim _{ h\rightarrow0 }{ \frac { 2\sin { \frac { \left( 3{ \left( x+h \right)  }^{ 2 }+x+h \right) -\left( 3{ x }^{ 2 }+x \right)  }{ 2 } \cos { \frac { \left( 3{ \left( x+h \right)  }^{ 2 }+x+h \right) +\left( 3{ x }^{ 2 }+x \right)  }{ 2 }  }  }  }{ h }  } \\ 
=&\lim _{ h\rightarrow 0 }{ \frac { 2\sin { \frac { h\left( 6x+3{ h }+1 \right)  }{ 2 } \cos { \frac { 6{ x }^{ 2 }+6xh+3{ h }^{ 2 }+2x+h }{ 2 }  }  }  }{ h }  }\\ 
=&\lim _{ h\rightarrow 0 }{ \frac { \sin { \frac { h\left( 6x+3{ h }+1 \right)  }{ 2 } \cos { \frac { 6{ x }^{ 2 }+6xh+3{ h }^{ 2 }+2x+h }{ 2 }  }  }  }{ \frac { h\left( 6x+3{ h }+1 \right)  }{ 2 }  }  } \left( 6x+3{ h }+1 \right) \\
 =&\lim _{ h\rightarrow 0 }{ \cos { \frac { 6{ x }^{ 2 }+6xh+3{ h }^{ 2 }+2x+h }{ 2 }  }  } \left( 6x+3{ h }+1 \right) \\
=&\cos { \left( \frac { 6{ x }^{ 2 }+2x }{ 2 }  \right) \left( 6x+1 \right)  } 
\end{eqnarray*}
