Differentiation of the 1st Degree.

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okpalandev

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May 30, 2024
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Hi , I am new to calculus,
After watching Eddy Woo video on YouTube, I stubbled on a premise and would like to `check` if it is correct. The question I pose:
The differentiation to the first degree of any subsequent variable is equal to the derivate rise/run gradient once.
For example:
```
x = 2x^3
dy/dx = 6x^2
```
when not derived again is the same as the rise/run. Which is the not the same as when you differentiate multiple time also know as the differentiation to the n^th degree 1
I am correct, if so can you show me how?
 
okpalandev said:
Hi , I am new to calculus,
After watching Eddy Woo video on YouTube, I stubbled on a premise and would like to `check` if it is correct. The question I pose:
The differentiation to the first degree of any subsequent variable is equal to the derivate rise/run gradient once.
For example:
```
x = 2x^3
dy/dx = 6x^2
```
when not derived again is the same as the rise/run. Which is the not the same as when you differentiate multiple time also know as the differentiation to the n^th degree 1
I am correct, if so can you show me how?
Click to expand...
It may help if you write the question in your native language so we can try to translate it for ourselves. Or maybe you need to give a link to the video so we can see what it is about.

But you seem just to be saying that the first derivative of a function gives the slope (gradient) of the tangent line, and higher-order derivatives (second derivative, third derivative, etc.) are different. What I don't understand is the last line: "show me how" ... how to do what?

Presumably the notation was meant to say

If y = 2x^3, then dy/dx = 6x^2​

which would be correct.
 
Unfortunately, mind reading technology are still being worked on.
1.Is the differentiation to the first degree of any subsequent variable is equal to the derivative rise/run gradient once?
2. Show how?

I was trying to show that differentiation was without doubt the inverse of intergration(antiderivate).
Fortunely, The Fundamental Theorem of Calculus states that the derivative of an integral with respect to the upper bound equals the integrand.
aready exist,it is just the proof.

Using the power rule:
The differetation to first degree is equal to derivate which can be calculated as:
```
f'(2x^3)=
d/dx(2x^3) = 6x^2
= tan(2x^3) for all x
```

The seond deriavate and the rationship to the degree of second differentation:
```
f''(2x^3) = f'(6x^2) = 12x
y= tan(2x^3) for all real number R
```
which i believe is not equavalent to tan(y)
 
Last edited:
okpalandev said:
Unfortunately, mind reading technology are still being worked on.
1.Is the differentiation to the first degree of any subsequent variable is equal to the derivative rise/run gradient once?
2. Show how?

I was trying to show that differentiation was without doubt the inverse of integration (antiderivative).
Fortunately, The Fundamental Theorem of Calculus states that the derivative of an integral with respect to the upper bound equals the integrand.
already exist, it is just the proof.

Using the power rule:
The differentiation to first degree is equal to derivative which can be calculated as:

f'(2x^3)=
d/dx(2x^3) = 6x^2
= tan(2x^3) for all x

The second derivative and the relationship to the degree of second differentiation:

f''(2x^3) = f'(6x^2) = 12x
y= tan(2x^3) for all real number R

which i believe is not equivalent to tan(y)
Click to expand...
The reason we need to read your mind is largely your imperfect English terminology; also, you have used the AsciiMath feature of the site (which I didn't even know still worked) imperfectly, so that expressions are not separated from one another. I've tried to fix some of that above.

I am not sure what you mean by "subsequent variable"; perhaps you mean "dependent variable".

Also, I am not sure whether you are using "first degree" to mean "the first order derivative", or perhaps the degree of a polynomial (the highest exponent).

Differentiation just means finding the derivative (yes, it is odd that the noun and verb don't match); so differentiation is equal to the derivative by definition; it doesn't need to be shown.

Where I have a problem is when you equate the derivative to tan(2x^3). What you are probably trying to say is that the derivative gives the slope of the curve at a given point; that's true. But that slope is the tangent of the angle the curve makes with the horizontal, not the tangent of the derivative itself.

I am also unsure what you are saying about the second derivative. But since you make the same mistake in expressing that, correcting the first error may be enough.

Finally, I don't see that anything you say is related to the antiderivative. The antiderivative is (again, by definition) the "inverse" of differentiation, taken in the right sense. It does not need to be "shown".
 
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