I have to differentiate with respect to $t_2$ on both sides of the following equation: $\dfrac{\partial c(x(t_2))}{\partial x} = p- t_2$. Can I do that in the following way?-
$\dfrac{\partial^2 c}{\partial x^2}*\dfrac{dx}{dt_2}=-1$
Please clarify. Thanks in advance.
I you can do that, since
$$f(x(t_2))=\frac{d c(x(t_2))}{d x}=c'(x(t_2))$$ $$\frac{d}{dt_2}\left\{\frac{dc(x(t_2))}{d x}\right\}= \frac{df(x(t_2))}{dt_2}=f'(x(t_2))\cdot x'(t_2)=\\c''(x(t_2))\cdot x'(t_2) =\frac{d^2 c(x(t_2))}{d x^2}\dfrac{dx(t_2)}{dt_2}= \frac{d^2 c}{d x^2}\dfrac{dx}{dt_2} $$