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For example, we can find the derivative of f(x) = e^(3x) this way:

f(x) = e^(3x)

f ' (x) = 3e^(3x)

f '' (x) = 9e^(3x)

f ''' (x) = 27e^(3x)

f^(4) (x) = 81e^(3x)

It looks like the pattern is that, each time we take the derivative, we multiply the function we had before by 3.

Multiplying f(x) by 3 fifty times gives (3^50)e^(3x).

So the 50th derivative of e^(3x) is (3^50)e^(3x).

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We find the 1st derivative , then the second derivative, and so on...

We could find a pattern of the n th derivative.

f(x) = a x^n a and n are constants.

if n < 50, f ⁵° = 0

if n = 50 , f ⁵° = 50! * a

if n > 50, f ⁵° = n (n-1)..(n-49) a x^(n-50)

=========================

f(x) = a Sin bx

f ' = a b cos bx

f '' = - a b² Sin bx = - b² f

f ³ = - a b³ Cos bx = - b² f '

f ⁴ = a b⁴ Sin bx = - b² f '' = b⁴ f

f ⁵ = a b⁵ Cos bx = - b² f ³ =

f ⁶ = - a b⁶ Sin bx = - b² f ⁴ = - b⁶ f

f ⁷ = - a b⁷ Cos bx = - b² f ⁵

f ⁸ = a b⁸ Sin bx = b⁸ f

so f ⁴⁸ = b⁴⁸ f

f ⁴⁹ = b⁴⁹ a Cos bx

f ⁵° = - b⁵° a Sin bx = - b⁵° f

==========================

f(x) = Log ax

f ' = a x⁻¹ f '' = - a x⁻² f ³ = 2 a x⁻³ f ⁴ = - 3! x⁻⁴

so f ⁵° = - 49! x⁻⁵°

We could find a pattern of the n th derivative.

f(x) = a x^n a and n are constants.

if n < 50, f ⁵° = 0

if n = 50 , f ⁵° = 50! * a

if n > 50, f ⁵° = n (n-1)..(n-49) a x^(n-50)

=========================

f(x) = a Sin bx

f ' = a b cos bx

f '' = - a b² Sin bx = - b² f

f ³ = - a b³ Cos bx = - b² f '

f ⁴ = a b⁴ Sin bx = - b² f '' = b⁴ f

f ⁵ = a b⁵ Cos bx = - b² f ³ =

f ⁶ = - a b⁶ Sin bx = - b² f ⁴ = - b⁶ f

f ⁷ = - a b⁷ Cos bx = - b² f ⁵

f ⁸ = a b⁸ Sin bx = b⁸ f

so f ⁴⁸ = b⁴⁸ f

f ⁴⁹ = b⁴⁹ a Cos bx

f ⁵° = - b⁵° a Sin bx = - b⁵° f

==========================

f(x) = Log ax

f ' = a x⁻¹ f '' = - a x⁻² f ³ = 2 a x⁻³ f ⁴ = - 3! x⁻⁴

so f ⁵° = - 49! x⁻⁵°