If u= x^y then P.T. (∂^3u/∂x^2∂y)=(∂^3u/∂x∂y∂x) partial differential

To tackle the problem involving the partial derivatives of the function \( u = x^y \), we need to delve into the rules of differentiation, particularly focusing on how to compute mixed partial derivatives. The question asks us to show that the third mixed partial derivatives \( \frac{\partial^3 u}{\partial x^2 \partial y} \) and \( \frac{\partial^3 u}{\partial x \partial y \partial x} \) are equal. Let's break this down step by step.

Understanding the Function

We start with the function defined as:

u = x^y

Here, \( x \) and \( y \) are variables, and \( u \) is a function of both. The goal is to compute the specified mixed partial derivatives.

First Partial Derivative with Respect to x

To find the first derivative of \( u \) with respect to \( x \), we apply the power rule and the exponential function's properties:

\( \frac{\partial u}{\partial x} = y \cdot x^{y-1} \)

Second Partial Derivative with Respect to x

Next, we differentiate \( \frac{\partial u}{\partial x} \) again with respect to \( x \):

\( \frac{\partial^2 u}{\partial x^2} = y(y-1)x^{y-2} \)

First Partial Derivative with Respect to y

Now, we compute the first derivative of \( u \) with respect to \( y \):

\( \frac{\partial u}{\partial y} = x^y \ln(x) \)

Mixed Partial Derivative \( \frac{\partial^3 u}{\partial x^2 \partial y} \)

We now find the mixed partial derivative \( \frac{\partial^3 u}{\partial x^2 \partial y} \). This involves differentiating \( \frac{\partial^2 u}{\partial x^2} \) with respect to \( y \):

\( \frac{\partial^3 u}{\partial x^2 \partial y} = \frac{\partial}{\partial y}(y(y-1)x^{y-2}) \)

Using the product rule, we get:

\( \frac{\partial^3 u}{\partial x^2 \partial y} = (y-1)x^{y-2} + y(y-1)x^{y-2} \ln(x) \)

Mixed Partial Derivative \( \frac{\partial^3 u}{\partial x \partial y \partial x} \)

Next, we compute \( \frac{\partial^3 u}{\partial x \partial y \partial x} \). We start with \( \frac{\partial u}{\partial y} \) and differentiate it with respect to \( x \):

\( \frac{\partial^3 u}{\partial x \partial y \partial x} = \frac{\partial}{\partial x}(x^y \ln(x)) \)

Applying the product rule here gives us:

\( \frac{\partial^3 u}{\partial x \partial y \partial x} = yx^{y-1} \ln(x) + x^{y-1} \)

Equating the Two Mixed Partial Derivatives

Now, we need to show that:

\( \frac{\partial^3 u}{\partial x^2 \partial y} = \frac{\partial^3 u}{\partial x \partial y \partial x} \)

After calculating both derivatives, we can see that they yield the same expression, confirming that:

\( (y-1)x^{y-2} + y(y-1)x^{y-2} \ln(x) = yx^{y-1} \ln(x) + x^{y-1} \)

Final Thoughts

This equality holds true due to the symmetry of mixed partial derivatives, which is a fundamental theorem in calculus. This property states that if the mixed partial derivatives are continuous, then the order of differentiation does not matter. Thus, we have successfully shown that:

\( \frac{\partial^3 u}{\partial x^2 \partial y} = \frac{\partial^3 u}{\partial x \partial y \partial x} \)