equality of $$ \frac{\partial^{2} t}{\partial x \partial y}=\frac{\partial^{2} t}{\partial y \partial x} $$ श्वार्ट्ज प्रमेय का कथन लिखकर सिद्ध कीजिये। की $$ \frac{\partial^{2} t}{\partial x \partial y}=\frac{\partial^{2} t}{\partial y \partial x} $$ State and prove Euler's thenram

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of mixed partial derivatives.

To prove the equality of mixed partial derivatives, we can use Euler's theorem of mixed partial derivatives, which states that if a function $$ t $$ is twice continuously differentiable with respect to $$ x $$ and $$ y $$, then the mixed partial derivatives are equal, i.e.,

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

Proof:

Let $$ t $$ be a function of two variables $$ x $$ and $$ y $$ such that $$ t $$ is twice continuously differentiable with respect to $$ x $$ and $$ y $$. We want to show that

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

To prove this, we can use the definition of the mixed partial derivative. The mixed partial derivative of $$ t $$ with respect to $$ x $$ and $$ y $$ is defined as

$$ \frac{\partial^{2} t}{\partial x \partial y} = \lim_{h \to 0} \frac{\frac{\partial t}{\partial y}(x+h, y) - \frac{\partial t}{\partial y}(x, y)}{h} $$

Similarly, the mixed partial derivative of $$ t $$ with respect to $$ y $$ and $$ x $$ is defined as

$$ \frac{\partial^{2} t}{\partial y \partial x} = \lim_{k \to 0} \frac{\frac{\partial t}{\partial x}(y+k, x) - \frac{\partial t}{\partial x}(y, x)}{k} $$

Now, let's consider the difference between the two mixed partial derivatives:

$$ \frac{\partial^{2} t}{\partial x \partial y} - \frac{\partial^{2} t}{\partial y \partial x} = \lim_{h \to 0} \frac{\frac{\partial t}{\partial y}(x+h, y) - \frac{\partial t}{\partial y}(x, y)}{h} - \lim_{k \to 0} \frac{\frac{\partial t}{\partial x}(y+k, x) - \frac{\partial t}{\partial x}(y, x)}{k} $$

By the definition of the derivative, we can rewrite the above expression as

$$ \frac{\partial^{2} t}{\partial x \partial y} - \frac{\partial^{2} t}{\partial y \partial x} = \lim_{h \to 0} \frac{\frac{\partial}{\partial x} \left( \frac{\partial t}{\partial y}(x+h, y) - \frac{\partial t}{\partial y}(x, y) \right)}{h} - \lim_{k \to 0} \frac{\frac{\partial}{\partial y} \left( \frac{\partial t}{\partial x}(y+k, x) - \frac{\partial t}{\partial x}(y, x) \right)}{k} $$

Since $$ t $$ is twice continuously differentiable, we can apply the chain rule to the derivatives inside the limits. This gives us

$$ \frac{\partial^{2} t}{\partial x \partial y} - \frac{\partial^{2} t}{\partial y \partial x} = \lim_{h \to 0} \frac{\frac{\partial}{\partial x} \left( \frac{\partial t}{\partial y}(x+h, y) - \frac{\partial t}{\partial y}(x, y) \right)}{h} - \lim_{k \to 0} \frac{\frac{\partial}{\partial y} \left( \frac{\partial t}{\partial x}(y+k, x) - \frac{\partial t}{\partial x}(y, x) \right)}{k} $$

$$ = \lim_{h \to 0} \frac{\frac{\partial}{\partial x} \left( \frac{\partial t}{\partial y}(x+h, y) - \frac{\partial t}{\partial y}(x, y) \right)}{h} - \lim_{k \to 0} \frac{\frac{\partial}{\partial y} \left( \frac{\partial t}{\partial x}(y+k, x) - \frac{\partial t}{\partial x}(y, x) \right)}{k} $$

\[ = \lim_{h \to 0} \frac{\frac{\partial}{\partial x} \left
Euler's theorem of mixed partial derivatives states that if a function $$ t $$ is twice continuously differentiable with respect to $$ x $$ and $$ y $$, then the mixed partial derivatives are equal, i.e., $$ \frac{\partial^{2} t}{\partial x \partial y} = \frac{\partial^{2} t}{\partial y \partial x} $$. The proof involves using the definition of the mixed partial derivative and applying the chain rule.

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