Part 1/6:

Understanding the Proof of Formula Six in Vector Function Differentiation

In advanced calculus and vector analysis, the rules for differentiating vector functions are fundamental. One such rule, often referred to as Formula Six, deals with the derivative of a vector function with a nested or composite function, akin to the chain rule in single-variable calculus. The goal of the exercise is to rigorously prove this formula, demonstrating how the derivative operates when the parameter of the vector function itself is a function of time.

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The Context of the Proof

Part 2/6:

The proof concerns a vector function ( \mathbf{u}(t) ), which has three components ( u_1(t) ), ( u_2(t) ), and ( u_3(t) ). Instead of being parameterized directly by ( t ), the function's argument is a more general function ( f(t) ). In essence, the task is to find:

[

\frac{d}{dt} \mathbf{u}(f(t))

]

and demonstrate that this derivative can be expressed in terms of the derivatives of ( \mathbf{u} ) evaluated at ( f(t) ), scaled by the derivative of ( f(t) ).

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Step-by-Step Breakdown of the Proof

Defining the Vector Function

Let:

[

\mathbf{u}(t) = \begin{bmatrix} u_1(t) \ u_2(t) \ u_3(t) \end{bmatrix}

]

which implies that:

[

\mathbf{u}(f(t)) = \begin{bmatrix} u_1(f(t)) \ u_2(f(t)) \ u_3(f(t)) \end{bmatrix}

]

Part 3/6:

The task is to differentiate this composition with respect to ( t ).

Differentiating Component-wise

Applying the chain rule component-wise yields:

[

\frac{d}{dt} \mathbf{u}(f(t)) = \begin{bmatrix} \frac{d}{dt} u_1(f(t)) \ \frac{d}{dt} u_2(f(t)) \ \frac{d}{dt} u_3(f(t)) \end{bmatrix}

]

Each component requires applying elementary differentiation rules, taking into account that ( u_j ) are functions of ( f(t) ), and ( f(t) ) itself depends on ( t ).

Applying the Chain Rule

For each component, the chain rule gives:

[

\frac{d}{dt} u_j(f(t)) = u_j'(f(t)) \cdot f'(t)

]

where ( u_j'(f(t)) ) is the derivative of ( u_j ) with respect to its own argument, evaluated at ( f(t) ), and ( f'(t) ) is the derivative of ( f(t) ).

Part 4/6:

This results in the entire derivative vector:

[

\frac{d}{dt} \mathbf{u}(f(t)) = \mathbf{u}'(f(t)) \cdot f'(t)

]

with ( \mathbf{u}'(f(t)) = [u_1'(f(t)), u_2'(f(t)), u_3'(f(t))]^\top ).

Vectorized Representation

To encapsulate the process efficiently, the proof expresses this as multiplication:

[

\frac{d}{dt} \mathbf{u}(f(t)) = \mathbf{u}'(f(t)) \cdot f'(t)

]

or, more explicitly,

[

\frac{d}{dt} \mathbf{u}(f(t)) = f'(t) \times \mathbf{u}'(f(t))

]

which confirms the applicability of the chain rule to vector functions with composite arguments.

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Final Result and Significance

Part 5/6:

The proof concludes that the derivative of a vector function with a composite parameter ( f(t) ) adheres to a straightforward rule: the derivative equals the derivative of the vector function evaluated at ( f(t) ) multiplied by the derivative of ( f(t) ).

This aligns with the fundamental chain rule in calculus and shows that its logic extends naturally to vector functions. Through careful component-wise differentiation and using the chain rule, the proof affirms the consistency of differentiation rules within vector calculus, providing a solid foundation for more complex analyses in physics, engineering, and applied mathematics.

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Conclusion

Part 6/6:

The exercise confirms the derivative rule for vector functions with a nested functional argument — an essential step in understanding how to differentiate more complex vector expressions commonly encountered in dynamical systems, physics, and computer graphics. By methodically applying the chain rule inside each component, the proof demonstrates that the operation maintains its integrity in higher dimensions, ensuring the tools mathematicians and scientists use are robust and reliable.