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,.

Answer
- The questioner was satisfied with and accepted the answer, or
- The answer was evaluated as being 100% correct by the judge.

-
Let me know if you need any clarifications.
-
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 ?
-
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.
-
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
-
That's correct.
-
- answered
- 720 views
- $5.00
Related Questions
- Find amplitude-frequency characteristic of a discrete finite signal using Z-transform
- Find the extrema of $f(x,y)=x$ subject to the constraint $x^2+2y^2=2$
- Evaluate $\int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx$
- Compound Interest question
- Connected rates of change calculus.
- Derivatives
- Use Stokes's Theorem to evaluate $\iint_S ( ∇ × F ) ⋅ d S$ on the given surface
- Calculus Integral Questins