Q. 75 - Calculus

Question

$$\begin{gathered} \mathrm{Let} f\left(x\right) \mathrm{be} \mathrm{a} \mathrm{differentiable} \mathrm{function} \mathrm{of} x, g\left(y\right) \mathrm{be} \mathrm{a} \mathrm{differentiable} \mathrm{function} \mathrm{of} y, \\ \mathrm{and} h(x, y)=f\left(x\right)+g\left(y\right). \mathrm{Prove} \mathrm{that} \frac{{\partial }^{2}h}{\partial x\partial y}=\frac{{\partial }^{2}h}{\partial y\partial x}. \end{gathered}$$

Step-by-Step Solution

Answer

$\frac{{\partial }^{2}h}{\partial x\partial y}=\frac{{\partial }^{2}h}{\partial y\partial x}$

Step 1. Given information

$$\begin{gathered} h(x, y)=f\left(x\right)+g\left(y\right) \\ \mathrm{Where}, f\left(x\right) is \mathrm{a} \mathrm{differentiable} \mathrm{function} \mathrm{of} x, g\left(y\right) is \mathrm{a} \mathrm{differentiable} \mathrm{function} \mathrm{of} y. \end{gathered}$$

Step 2. Proof of given partial derivative

$$\begin{gathered} \mathrm{LHS}=\frac{{\partial }^{2}h}{\partial x\partial y} \\ =\frac{\partial }{\partial x}\left(\frac{\partial \left(f\right(x)+g(y\left)\right)}{\partial y}\right) \\ =\frac{\partial (g\text{'}(y\left)\right)}{\partial x} \\ =0 \\ \mathrm{RHS}=\frac{{\partial }^{2}h}{\partial y\partial x} \\ =\frac{\partial }{\partial y}\left(\frac{\partial \left(f\right(x)+g(y\left)\right)}{\partial x}\right) \\ =\frac{\partial (f\text{'}(x\left)\right)}{\partial y} \\ =0 \\ \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$