Given a function \displaystyle{f(x)=x^k}
where k
is an integer greater other than 0, to calculate the n-th derivative of f(x)
you can use the following formula:\displaystyle{f^{(n)}(x) = \begin{cases} \left(\prod_{j=0}^{n-1} k-j\right)x^{k-n} & 0 < n < k-1\\((n+1)!)x & n = k - 1\\n! & n = k \\ 0 & n > k \end{cases}}
How do we get to this conclusion?
By observation using the product rule of the derivative of x^n
. By using the product rule on f(x)
we get:\displaystyle{f'(x)=kx^{k-1}}
\displaystyle{f''(x)=k(k-1)x^{k-2}}
\displaystyle{f^{(3)}(x)=k(k-1)(k-2)x^{k-3}}
...\displaystyle{f^{(n)}=k(k-1)(k-2)...(k-(n-1))x^{k-n}}
The coefficient of f^{(n)}(x)
can be expressed as\displaystyle{(k-0)(k-1)(k-2)...(k-(n-1))}
Which can be simplified to\displaystyle{\prod_{j=0}^{n-1} k-j}
And replacing this in the original function, we get\displaystyle{f^{(n)}=\left(\prod_{j=0}^{n-1} k-j\right)x^{k-n}}
Proof by induction
First, let's try any specific case to see if the formula works. Let's say n = 4
and k = 6
. The formula is:\displaystyle{f^{(4)}(x)=\left(\prod_{j=0}^{4-1} 6-j\right)x^{6-4} =}
\displaystyle{= [(6-0)(6-1)(6-2)(6-3)]x^{6-4} = (6\cdot 5 \cdot 4 \cdot 3)x² = 360x²}
A quick check by using the product rule on f(x)
four times show us that this is correct. Now we can asume that for any positive integers n and k, this formula works. If this is correct for an integer n, then it must also be true for any integer n + 1.
Let's add an extra term to f^{(n)}(x)
and see what happens:\displaystyle{f^{(n+1)}=k(k-1)(k-2)...(k-(n-1))(k-n)x^{k-(n+1)}}
Now let's plug n+1
to the other formula:\displaystyle{f^{(n+1)}=\left(\prod_{j=0}^{n+1-1} k-j\right)x^{k-(n+1)} = \left(\prod_{j=0}^{n} k-j\right)x^{k-(n+1)}}
We can extract the last term from the product:\displaystyle{(k-n)\left(\prod_{j=0}^{n-1} k-j\right)x^{k-(n+1)}}
Which, if you expand the product, is exactly the same as f^{(n+1)}(x)
as we calculated earlier. This proves that this part of the function is correct and works for any positive integer n.
The rest of the segments
This part of the function works for any positive integers 0 < n < k - 1
. The reason being the way the product rule works. We can see how it works from observation. Asume k = 6
, then:f^{(1)}(x) = 6x⁵
, f^{(2)}(x) = 30x⁴
, f^{(3)}(x) = 120x³
, f^{(4)}(x) = 360x²
, f^{(5)}(x) = 720x
From the sixth derivative onward, the product rule no longer applies.f^{(6)}(x) = 720
, f^{(7)}(x) = 0
You will notice that exactly at n = k-1
, f^{(k-1)}(x)=((n+1)!)x
. If we go to the general formula for f^{(n)}(x)
and replace n = k - 1
, we get:\displaystyle{f^{(k-1)}(x)=k(k-1)(k-2)...(k-k+1)x^{k-k+1}}
\displaystyle{f^{(k-1)}(x)=(k(k-1)(k-2)... \cdot 3 \cdot 2 \cdot 1 )x^{1}}
\displaystyle{f^{(k-1)}(x)=(k!)x}
And since n = k -1
, or k=n+1
, we conclude that\displaystyle{f^{(n)}(x)=((n+1)!)x}
The rest of the function is just the application of the derivative rules. Since x
has an implied power of 1, the power rule does not apply any longer and it's derivative is 1. And after that, the derivative of a constant value is always 0.
Posts
-
Given a function \displaystyle{f(x)=x^k} where k is an integer greater other than 0, to calculate the n-th derivative of f(x) you can use the following formula:\displaystyle{f^{(n)}(x) = \begin{cases} \left(\prod_{j=0}^{n-1} k-j\right)x^{k-n} & 0 < n <...