Nth Derivative of a Polynomial (part 1)

In calculus, the derivative of a polynomial is:axn=naxn1 { 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.

  1. If n<bn<b , then the nth derivative of the polynomial would be

dndxnaxb=a(bn+1)(bn+2)....(bn+n)xbn\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 4x5{ 4x }^{ 5 }?

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

d3dx34x5=4(53+1)(53+2)(53+3)x53\frac { { d }^{ 3 } }{ { dx }^{ 3 } } { 4x }^{ 5 }=4(5-3+1)(5-3+2)(5-3+3){ x }^{ 5-3 }

d3dx34x5=4(3)(4)(5)x53\frac { { d }^{ 3 } }{ { dx }^{ 3 } } { 4x }^{ 5 }=4(3)(4)(5){ x }^{ 5-3 }

d3dx34x5=240x2\frac { { d }^{ 3 } }{ { dx }^{ 3 } } { 4x }^{ 5 }={ 240x }^{ 2 }

Note by Merzel Mark Guilaran
4 years, 9 months ago

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

Tapas Mazumdar - 1 year, 6 months ago

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