Find the first four non-zero terms of the Maclaurin series for
$x^3 \sin(x^2)$.
I found the derivatives and got as following:
first derivative = $2x^3 \cos(x^2)$
second derivative = $-12x^3 \sin(x^2)$
third derivative = $-72x^3 \cos(x^2)$
when I used formula, I am substituting $x=0$ and getting all the first four terms as 0.
Please help to solve this.
Best Answer
Notice to differentiate $f(x) = x^3 \sin (x^2)$ you must use both the chain rule and the product rule. It follows that
$$ f'(x) = 3 x^ 2 \sin (x^2) + 2 x^4 \cos (x^2) $$
But, notice that since $\sin x = \sum \frac{ (-1)^n x^{2n+1 } }{(2n+1)!} $, then
$$ x^3 \sin (x^2) = x^3 \sum_{n \geq 0} \frac{ (-1)^n x^{4n+2} }{(2n+1)!} = \sum_{n \geq 0 } \frac{ (-1)^n x^{4n+5}}{(2n+1)!} $$
Thus, for example, the first and second term are $$ x^5, \frac{ - x^9 }{3!} $$
and so on.