Directional Derivatives .... Apostol, Section 12.2, Example 4 ....

[Math Amateur]

I am reading Tom M Apostol's book "Mathematical Analysis" (Second Edition) ...I am focused on Chapter 12: Multivariable Differential Calculus ... and in particular on Section 12.2: The Directional Derivative ... ...I need help with part of Example 4, Section 12.2 ...Section 12.2, including the Examples, reads as follows:
View attachment 8498
View attachment 8499
In Example 4 above, we read the following:

"More generally, \(\displaystyle F'(t) = f'(c + tu; u)\) if either derivative exists."Can someone help me to show, formally and rigorously, that \(\displaystyle F'(t) = f'(c + tu; u)\) ... ...Hope someone can help ...

Peter

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It may help MHB readers of the above post to have access to Apostol's definition of the derivative of a function of one real variable ... so I am providing the same as follows:
View attachment 8500

Hope that helps ...

Peter

 

[Chris L T521]

I am reading Tom M Apostol's book "Mathematical Analysis" (Second Edition) ...I am focused on Chapter 12: Multivariable Differential Calculus ... and in particular on Section 12.2: The Directional Derivative ... ...I need help with part of Example 4, Section 12.2 ...Section 12.2, including the Examples, reads as follows:In Example 4 above, we read the following:

"More generally, \(\displaystyle F'(t) = f'(c + tu; u)\) if either derivative exists."Can someone help me to show, formally and rigorously, that \(\displaystyle F'(t) = f'(c + tu; u)\) ... ...Hope someone can help ...

Peter

By definition of the derivative,

\[\begin{aligned}\mathbf{F}^{\prime}(t) &= \lim_{h\to 0}\frac{\mathbf{F}(t+h)-\mathbf{F}(t)}{h}\\ &= \lim_{h\to 0}\frac{\mathbf{f}(\mathbf{c}+(t+h)\mathbf{u})-\mathbf{f}(\mathbf{c}+t\mathbf{u})}{h}\\ &= \lim_{h\to 0}\frac{\mathbf{f}(\mathbf{c}+t\mathbf{u}+h\mathbf{u})-\mathbf{f}(\mathbf{c}+t\mathbf{u})}{h}\\ &= \mathbf{f}^{\prime}(\mathbf{c}+t\mathbf{u};\mathbf{u})\quad\text{by definition of directional derivative}\end{aligned}\]

Setting $t=0$ yields $\mathbf{F}^{\prime}(0)=\mathbf{f}^{\prime}(\mathbf{c};\mathbf{u})$.

I hope this clarifies things!

EDIT: I used a different definition of the derivative...but it shouldn't be too hard to see why this is still true.

 

[Math Amateur]

By definition of the derivative,

\[\begin{aligned}\mathbf{F}^{\prime}(t) &= \lim_{h\to 0}\frac{\mathbf{F}(t+h)-\mathbf{F}(t)}{h}\\ &= \lim_{h\to 0}\frac{\mathbf{f}(\mathbf{c}+(t+h)\mathbf{u})-\mathbf{f}(\mathbf{c}+t\mathbf{u})}{h}\\ &= \lim_{h\to 0}\frac{\mathbf{f}(\mathbf{c}+t\mathbf{u}+h\mathbf{u})-\mathbf{f}(\mathbf{c}+t\mathbf{u})}{h}\\ &= \mathbf{f}^{\prime}(\mathbf{c}+t\mathbf{u};\mathbf{u})\quad\text{by definition of directional derivative}\end{aligned}\]

Setting $t=0$ yields $\mathbf{F}^{\prime}(0)=\mathbf{f}^{\prime}(\mathbf{c};\mathbf{u})$.

I hope this clarifies things!

EDIT: I used a different definition of the derivative...but it shouldn't be too hard to see why this is still true.

Hi Chris ...

Thanks for the help ...

Very clear indeed ... and most helpful ...

Peter