Nth Derivative of a Polynomial (part 1)

In calculus, the derivative of a polynomial is: ${ ax }^{ n }={ nax }^{ n-1 }$ . But how could we know the 2nd, 3rd or even the 100th derivative of a given polynomial expression. I formulated new formulas that would ensure that the time in finding the number of derivative would be less in calculating the higher derivatives. These are the following conditions:

Let's assume that n is the no. of differentiation or derivatives that a polynomial must derive and b be the exponents of each term in a given polynomial.

If $n<b$ , then the nth derivative of the polynomial would be

$\frac { { d }^{ n } }{ { dx }^{ n } } { ax }^{ b }=a(b-n+1)(b-n+2)....(b-n+n){ x }^{ b-n }$ .

Example: Find the third derivative of ${ 4x }^{ 5 }$ ?

Solution: Since the condition is $n<b$ then let's use the formula stated in first condition. Let n be 3 and b be 5.Therefore, the solution would be:

$\frac { { d }^{ 3 } }{ { dx }^{ 3 } } { 4x }^{ 5 }=4(5-3+1)(5-3+2)(5-3+3){ x }^{ 5-3 }$

$\frac { { d }^{ 3 } }{ { dx }^{ 3 } } { 4x }^{ 5 }=4(3)(4)(5){ x }^{ 5-3 }$

$\frac { { d }^{ 3 } }{ { dx }^{ 3 } } { 4x }^{ 5 }={ 240x }^{ 2 }$

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Yes in general $\dfrac{d^n}{dx^n} (ax^r) = an! {\dbinom{r}{n}} x^{r-n}$ . Good insight by the way sir.

Tapas Mazumdar - 3 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$