# Use taylor series to evaluate f^11 (0) for f(x) = x^3 cos(x^2) ?

Aug 7, 2018

The answer is $990$.

#### Explanation:

We can rewrite as follows:

f(x) = x^3(1 - x^2/2 + x^4/(4!)- x^6/(6!) + x^8/(8!) - x^10/(10!) + ...)

f(x) = x^3 - x^5/2 + x^7/(4!) - x^9/(6!) + x^11/(8!) - x^13/(10!)

Now note that we don't need any further terms because when we compute the 11th derivative all the terms except that of original degree $11$ will cancel to $0$.

f^11(0) = 11!/(8!) x^0 = (11!)/(8!) = 990

Hopefully this helps!