Question

Find the derivative using first principles? : `x^n`

Answer 1

We can do this via the use of first principles...

Explanation:

We must first derive the idea of a derivative;

using this idea we must use this for `f(x) = x^n`

to yields;

`lim_(h->0) ((x+h)^n - x^n)/(h)`

Now we must cosnider the expansion of `(x+h)^n`

We use `(alpha + beta)^n` = `alpha^n + (nC1) alpha^(n-1)beta + ... + beta^n`

So hence `(x+h)^n = x^n + (nC1)x^(n-1)h + ...`

hence the limit becomes; `lim_(h->0) (nC1)x^(n-1) + (nC2)x^(n-2)h + ...`

`= (nC1)x^(n-1)`

and we know `nC1 = n`

So hence yields;

`d/(dx) ( x^n) = nx^(n-1)`

Answer 2

See below.

Explanation:

Using the power rule:

`d/dx(x^n)=nx^(n-1)`

Example:

`d/dx(x^4)=(4)x^(4-1)=4x^3`

Answer 3

Please see below.

Explanation:

Verify (by multiplication) that for positive integer `n`,

`x^n-t^n=(x-t)(x^(n-1)+x^(n-2)t + x^(n-3)t^2 + * * * +xt^(n-2)+t^(n-1))`

`d/dx(x^n) = lim_(trarrx)(x^n-t^n)/(x-t)`

`= lim_(trarrx)((x-t)(x^(n-1)+x^(n-2)t + x^(n-3)t^2 + * * * +xt^(n-2)+t^(n-1)))/(x-t)`

`= lim_(trarrx)(x^(n-1)+x^(n-2)t + x^(n-3)t^2 + * * * +xt^(n-2)+t^(n-1))`

There are `n` terms, each with limit `x^(n-1)`, so the limit is

`d/dx(x^n) = nx^(n-1)` for positive integer `n`.

Answer 4

`d/dx x^n= nx^(n-1)`

Explanation:

Using the limit definition of the derivative then if:

`y = f(x) = x^n`

Then we have:

`dy/dx = lim_(h rarr 0) (f(x+h) - f(x))/h`
`\ \ \ \ \ = lim_(h rarr 0) ((x+h)^n - x^n)/h`

Then using the Binomial Theorem, we can expand to get:

`dy/dx = lim_(h rarr 0) ({x^n+nx^(n-1)h+...+h^n} - x^n)/h`
`\ \ \ \ \ = lim_(h rarr 0) (nx^(n-1)h+...+h^n)/h`
`\ \ \ \ \ = lim_(h rarr 0) nx^(n-1)+...+h^(n-1)`

Note that all the terms on the right, apart from the first, contain the term `h`. In the limit as `h` tends to zero, all these become zero.

`:. dy/dx = nx^(n-1)`