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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## Comments

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

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