Power rule

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shebi.zac

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Greetings... please see below attached photo .. How can delta x be equated to zero ( towards the last step) as it is only approaching to zero ? What is the logic behind ?If it can be equated to zero... could it have been equated to zero at the first step it self??
Please help BFAA5C80-6FF9-4106-AE97-252FC26C2934.pngBFAA5C80-6FF9-4106-AE97-252FC26C2934.png
 
Because if you have done it in the first step, you will divide over zero [MATH]\frac{(x + 0)^n - x^n}{0}[/MATH] which is not allowed.
 
Yes ... Thank you sir... But May I know what is the reason/logic behind when Delta x is equated to zero as its seemingly only approaching zero in the last step...
 
As Harry the cat said, delta x is never "equated to 0". you take the limit as delta x goes to 0. Do you understand the "limit" concept?
 
I prefer to do all the algebra BEFORE taking a limit.

[MATH]f(x) = x^n,\ n \text { is an integer} > 1.[/MATH]
[MATH]\Delta x \ne 0.[/MATH]
[MATH]f(x + \Delta x) = (x + \Delta x)^n = \sum_{j=0}^n \dbinom{n}{j} x^{(n - j)} \Delta x^j = x^n + nx^{(n-1)} \Delta x + \Delta x^2 \left ( \sum_{j=2}^n \dbinom{n}{j} x^{(n-j)} \Delta x^{(j-2)} \right ).[/MATH]
[MATH]\therefore f(x + \Delta x) - f(x) = nx^{(n-1)} \Delta x + \Delta x^2 \left ( \sum_{j=2}^n \dbinom{n}{j} x^{(n-j)} \Delta x^{(j-2)} \right ) \implies[/MATH]
[MATH]\dfrac{f(x + \Delta x) - f(x)}{\Delta x} = nx^{(n-1)} + \Delta x \left ( \sum_{j=2}^n \dbinom{n}{j} x^{(n-j)} \Delta x^{(j-2)} \right ) [/MATH]
Now take the limit. As delta x goes toward zero, is it not intuitive that the term multiplied by delta x also goes toward zero leaving the other term unaffected?
 
JeffM said:
I prefer to do all the algebra BEFORE taking a limit.

[MATH]f(x) = x^n,\ n \text { is an integer} > 1.[/MATH]
[MATH]\Delta x \ne 0.[/MATH]
[MATH]f(x + \Delta x) = (x + \Delta x)^n = \sum_{j=0}^n \dbinom{n}{j} x^{(n - j)} \Delta x^j = x^n + nx^{(n-1)} \Delta x + \Delta x^2 \left ( \sum_{j=2}^n \dbinom{n}{j} x^{(n-j)} \Delta x^{(j-2)} \right ).[/MATH]
[MATH]\therefore f(x + \Delta x) - f(x) = nx^{(n-1)} \Delta x + \Delta x^2 \left ( \sum_{j=2}^n \dbinom{n}{j} x^{(n-j)} \Delta x^{(j-2)} \right ) \implies[/MATH]
[MATH]\dfrac{f(x + \Delta x) - f(x)}{\Delta x} = nx^{(n-1)} + \Delta x \left ( \sum_{j=2}^n \dbinom{n}{j} x^{(n-j)} \Delta x^{(j-2)} \right ) [/MATH]
Now take the limit. As delta x goes toward zero, is it not intuitive that the term multiplied by delta x also goes toward zero leaving the other term unaffected?
Click to expand...
Thank you for the elaborate explanation
 
For fixed x:
[MATH] \text{ }\\ \dfrac{(x+\Delta x)^n-x^n}{\Delta x} =nx^{n-1} + \Delta x \hspace1ex p(\Delta x) \hspace4ex p(\Delta x) \text{ is a polynomial in }\Delta x \hspace1ex (x \text{ fixed).}\\ \text{ }\\ \text{Now when }|\Delta x|<1 \text{ say, } \hspace1ex |p(\Delta x)| \text{ is bounded by a constant K (dependent on the fixed x)}\\ \text{ }\\ \therefore |\Delta x \; p(\Delta x)| ≤ |\Delta x| K \\ \text{ }\\ \therefore \lim \limits_{\Delta x \to 0} \Delta x \; p(x) = 0\\ \text{ }\\ \therefore \lim \limits_{\Delta x \to 0} \dfrac{(x+\Delta x)^n-x^n}{\Delta x} = nx^{n-1}\\ [/MATH]
 
The fact that the derivative of \(\displaystyle x^n\) is \(\displaystyle nx^{n-1}\) can also be proved by induction on n. When n= 1, this is \(\displaystyle x^1= x\). That's linear so the derivative is just the constant slope, \(\displaystyle 1= 1(x^0)= 1(x^{1-1}\). Assume that, for n= k, the derivative of \(\displaystyle x^k]\) is \(\displaystyle kx^{k-1}\) Then the derivative of \(\displaystyle x^{k+1}= x(x^k)\) is, by the product rule, \(\displaystyle (1)(x^k)+ (x)(kx^{k-1}= x^k+kx^k= (k+1)x^k= nx^{n-1}\).
 
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