If u, v and w are functions of x, then show that d(u.v.w)/dx = (du/dx.v.w + u.dv/dx.w + u.v.dw/dx) in two ways - first by repeated application of product rule, second by logarithmic differentiation.

To demonstrate that the derivative of the product of three functions \( u \), \( v \), and \( w \) with respect to \( x \) can be expressed as \( \frac{d(uvw)}{dx} = \left(\frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx}\right) \), we will use two different methods: the product rule and logarithmic differentiation.

Method 1: Repeated Application of the Product Rule

The product rule states that the derivative of a product of two functions \( f \) and \( g \) is given by:

\( \frac{d(fg)}{dx} = f\frac{dg}{dx} + g\frac{df}{dx} \)

We can apply this rule step by step for three functions:

1. First, consider \( u \) and \( v \). The derivative of their product is:

\( \frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \)

2. Next, we treat \( uv \) as a single function and multiply it by \( w \). The derivative becomes:

\( \frac{d(uvw)}{dx} = \frac{d(uv)}{dx} \cdot w + uv\frac{dw}{dx} \)

3. Substituting the first derivative into this equation gives:

\( \frac{d(uvw)}{dx} = \left(u\frac{dv}{dx} + v\frac{du}{dx}\right)w + uv\frac{dw}{dx} \)

4. Finally, we can rearrange this to match the desired form:

\( \frac{d(uvw)}{dx} = u\frac{dv}{dx}w + v\frac{du}{dx}w + uv\frac{dw}{dx} \)

Method 2: Logarithmic Differentiation

Logarithmic differentiation involves taking the logarithm of both sides of an equation. For the product \( y = uvw \), we start with:

\( \ln(y) = \ln(u) + \ln(v) + \ln(w) \)

Next, we differentiate both sides with respect to \( x \):

\( \frac{1}{y}\frac{dy}{dx} = \frac{1}{u}\frac{du}{dx} + \frac{1}{v}\frac{dv}{dx} + \frac{1}{w}\frac{dw}{dx} \)

Multiplying through by \( y \) (which is \( uvw \)) gives:

\( \frac{dy}{dx} = y\left(\frac{1}{u}\frac{du}{dx} + \frac{1}{v}\frac{dv}{dx} + \frac{1}{w}\frac{dw}{dx}\right) \)

Substituting back \( y = uvw \) results in:

\( \frac{dy}{dx} = uvw\left(\frac{1}{u}\frac{du}{dx} + \frac{1}{v}\frac{dv}{dx} + \frac{1}{w}\frac{dw}{dx}\right) \)

Distributing \( uvw \) leads to:

\( \frac{dy}{dx} = v w \frac{du}{dx} + u w \frac{dv}{dx} + u v \frac{dw}{dx} \)

This matches the desired expression, confirming that both methods yield the same result.