IS partial derivative sqared is like the second partial derivative?

Question

is $$(\frac{\partial P}{\partial T})^2=\frac{\partial^2P}{\partial T ^2}$$ holds? thank you

Answer 1

That equality does not hold; their perceived similarity comes from notation alone. The $^2$ in the RHS refers to the fact that it's a second derivative.

$$(\frac{\partial P}{\partial T})^2=\frac{\partial P}{\partial T}\cdot\frac{\partial P}{\partial T}$$ and

$$\frac{\partial^2P}{\partial T ^2} = \frac{\partial}{\partial T}\bigg[\frac{\partial P}{\partial T}\bigg]$$

Paul's example is a good one: Let $P = T + c$. Then

$$(\frac{\partial P}{\partial T})^2=\frac{\partial P}{\partial T}\cdot\frac{\partial P}{\partial T} = (1)(1) = 1$$ and

$$\frac{\partial^2P}{\partial T ^2} = \frac{\partial}{\partial T}\bigg[\frac{\partial P}{\partial T}\bigg] = \frac{\partial}{\partial T}\bigg[1\bigg] = 0$$

Answer 2

No. Counterexample: $P(T)=T$ $$ \left(\frac{\partial P}{\partial T}\right)^2=1\neq 0=\frac{\partial^2P}{\partial T^2}. $$

Answer 3

No. $\dfrac{\partial^2P}{\partial T^2}\;$ is denoted with the ewponent at a different place in the numerator and the denominator because it is a contraction for $$\dfrac{\partial }{\partial T}\biggl(\dfrac{\partial }{\partial T}\biggr)(P). $$

For functions of a single variable, equality would mean that $$f''=(f')^2, $$ which is clearly absurd: a polynomial of degree $n$ would have a second derivative of degree both $n-2$ and $2(n-1)$.

Answer 4

No. For example, let $P(T,V)=aT+bV$... Then $\frac {\partial P} {\partial T}=A$, so $(\frac {\partial P} {\partial T})^2=a^2$, but $\frac {\partial ^2 P}{\partial T ^2}=0$...