How to prove that $T \frac{d^2 T}{d x^2} = T f \frac{df}{dT}$

Question

Considering that:

\begin{equation} f(T) = \frac{dT}{dx} \end{equation}

and:

\begin{equation} T \frac{d^2 T}{d x^2} = 0 \end{equation}

\begin{equation} T f \frac{df}{dT} = 0 \end{equation} how to prove that

\begin{equation} T \frac{d^2 T}{d x^2} = T f \frac{df}{dT} \end{equation}

score 3 · Answer 1 · answered Aug 31 '22 at 22:12

3

The claim $T \frac{d^2 T}{d x^2} = T f \frac{df}{dT}$ simply follows from the fact that both sides of the equation are equal to $0$, according to the information you provided.

answered Aug 31 '22 at 22:12

Andreas Tsevas

2,520

score 2 · Accepted Answer · answered Aug 31 '22 at 22:45

According to the properties of differentials (assuming we don’t yet know that given expressions are both $0$):$$Tf\frac{df}{dT} = T\cdot \frac{dT}{dx}\cdot\frac{d\left(\frac{dT}{dx}\right)}{dT}=T\cdot \frac{dT}{dx}\frac{\frac{d^2T}{dx}}{dT} = T \cdot \frac{dT}{dT}\cdot\frac{d^2T}{dx dx} = T\frac{d^2T}{dx^2}$$

How to prove that $T \frac{d^2 T}{d x^2} = T f \frac{df}{dT}$

2 Answers2