# Thread: Caculate Dervitiv Using Exponential Series?

1. ## Caculate Dervitiv Using Exponential Series?

Let f(x) = e^x^3
Calculate the 9th derivative f (9)(0) using the exponential
series and Taylor's formula.
Kay am not sure if idid this correect but the series for e^x is

infinte series n=0 to infinty x^n/n! do i just plug in the equation and solve i know its zero but how??

2. Originally Posted by zangestu888
Let f(x) = e^x^3
Calculate the 9th derivative f (9)(0) using the exponential
series and Taylor's formula.
Kay am not sure if idid this correect but the series for e^x is

infinte series n=0 to infinty x^n/n! do i just plug in the equation and solve i know its zero but how??
zero, huh?

$\displaystyle e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

$\displaystyle e^{x^3} = 1 + x^3 + \frac{x^6}{2!} + \frac{x^9}{3!} + ...$
...

after 9 derivatives, the third term will be a constant equal to

$\displaystyle \frac{9!}{3!} = 60480$.

every subsequent term will have factors of x (which will become 0 when evaluating the 9th derivative at x = 0)

3. Can you please explain what you did? is thier like a shorcut to finding it using the exponential series and Taylor formula