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

4 Answers
Dec 3, 2017

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

Explanation:

We must first derive the idea of a derivative;

enter image source here

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) #

Dec 3, 2017

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#

Dec 4, 2017

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#.

Dec 4, 2017

# 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) #