maths help (1 Viewer)

[fullonoob]

if f(x) = x^n, find :
(n)
f (x)

the n is like f'(x)
Also
(k)
f (x)
explain please

 

[Trebla]

I assume you mean the n-th deriviative of f(x) = xⁿ?
f'(x) = nx^{n - 1}
f''(x) = n(n - 1)x^{n - 2}
f⁽³⁾(x) = n(n - 1)(n - 2)x^{n - 3}
.
.
.
f^(k)(x) = n(n - 1)(n - 2)......(n - k + 1)x^{n - k}
.
.
.
f⁽ⁿ⁾(x) = n(n - 1)(n - 2)......3.2.x^{n - n}
= n!

Notice the pattern of correlation between the last bracket and the index of x with the order of the derivative.

 

[fullonoob]

I assume you mean the n-th deriviative of f(x) = xⁿ?
f'(x) = nx^{n - 1}
f''(x) = n(n - 1)x^{n - 2}
f⁽³⁾(x) = n(n - 1)(n - 2)x^{n - 3}
.
.
.
f^(k)(x) = n(n - 1)(n - 2)......(n - k + 1)x^{n - k}
.
.
.
f⁽ⁿ⁾(x) = n(n - 1)(n - 2)......3.2.x^{n - n}
= n!

Notice the pattern of correlation between the last bracket and the index of x with the order of the derivative.

the ans for my second question is:
n(n-1)(n-2)...(n-k+1)x^n-k = (n!/(n-k)!) x^n-k
explain please

 

[fullonoob]

ALSO
Show that kxk!=(k+1)!-k!

 

[tommykins]

is that k multiplied by k! ?

k.k! = k!(k+1-1) = k!(k+1) - k! = (k+1)! - k!