Explain proof of directional derivative

I will try to mark the exact thing I don't understand, it comes from this thread https://matchmaticians.com/questions/u6me4r#answers :
What I don't understand is when we have $$g(h)=f(x_0+ha,y_0+hb)$$
That means that when 'h'(independent variable) changes in g(h), it will produce the same change in 'h' of f(x0​+ha,y0​+hb).
Derivative formula says that:
$$g'(x) = \lim_{{h \to 0}}\frac{g(x+h)-g(x)}{h}$$
so just changing variable name I receive
$$g'(h) = \lim_{{\Delta{h} \to 0}}\frac{g(h+\Delta{h})-g(h)}{\Delta{h}}$$
And when I want to find it's derivative at 0 I receive:
$$g'(0) = \lim_{{\Delta{h} \to 0}}\frac{g(0+\Delta{h})-g(0)}{\Delta{h}} = \lim_{{\Delta{h} \to 0}}\frac{g(\Delta{h})-g(0)}{\Delta{h}}$$
So the question is since $$g(h)=f(x_0+ha,y_0+hb)$$
Why can I replace $$g(\Delta{h})  by  f(x_0+ha,y_0+hb)$$
Since by replacing it I will receive not $$f(x_0+ha,y_0+hb)$$ but $$f(x_0+\Delta{h}a,y_0+\Delta{h}b)$$
P.S. I understand how directional derivatives work, I don't want you to explain it, I just want the explanation of this particular case please, feel free to ask for more money,.

Calculus
Babaduras Babaduras
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Erdos Erdos
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  • Erdos Erdos
    0

    Let me know if you need any clarifications.

  • Babaduras Babaduras
    0

    yeah, I think I was confusing delta(h) and (h) all the time. so correct case for derivative of g at 0 is g'(0) = limit of (g(delta(h) - g(0))/delta(h) with delta(h) approaching zero, replaceing g(h) with f(x0​+ha,y0​+hb) will give me limit of (f(x0​+delta(h)a,y0​+(delta(h)b) - f(x0​,y0​))/delta(h) with delta(h) approaching zero ?

    • Erdos Erdos
      0

      Replacing g(h) with f(x0​+ha,y0​+hb) will give me limit of (f(x0​+ha,y0​+hb) - f(x0​,y0​))/h with h approaching zero. You statement "replaceing g(h) with f(x0​+ha,y0​+hb) will give me limit of (f(x0​+delta(h)a,y0​+(delta(h)b) - f(x0​,y0​))/delta(h) with delta(h) approaching zero" has the same problem. Remember, you either use h or \Delta h. It does not make sense to use both in one computation.

    • Babaduras Babaduras
      0

      yeah, I got it, I mean that since g(h) = f(x0​+ha,y0​+hb), then replacing h with delta(h) will give me imit of (f(x0​+delta(h)a,y0​+(delta(h)b) - f(x0​,y0​))/delta(h) with delta(h) approaching zero

    • Erdos Erdos
      0

      That's correct.

The answer is accepted.
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