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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.

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    $\begingroup$ Your derivatives are very incorrect. Do you know the product rule? And the chain rule? $\endgroup$ – T. Bongers Apr 3 '18 at 23:35
  • $\begingroup$ Also, to save yourself some trouble, you don't have to take any derivatives if you know the Taylor expansion for $\sin(x)$. $\endgroup$ – Clayton Apr 3 '18 at 23:36
  • $\begingroup$ But, Maclaurin series need to have derivatives. $\endgroup$ – raasilinin Apr 3 '18 at 23:38
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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.

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Hint:

The derivative of a product is not the product of the derivatives of the factors. Instead, $(fg)'=f'g+fg'$, ad more generally, for the derivative of order $n$, we have Leibniz' formula: $$(fg)^{(n)}(x)=\sum_{k=0}^n \binom nk f^{(n-k)}(x)\,g^{(k)}(x).$$

This being said, you can obtain The Maclautin series from the Maclaurin series for $\sin u$, substituting $u=x^2$, and multiplying by $x ^3$.

Indeed, the first $4$ terms for the sine are $$\sin u=u-\frac{u^3}6+\frac{u^5}{120}-\frac{u^7}{5040}+O(u^8),$$ so we obtain \begin{align} x^3\sin x^2&=x^3\Bigl(x^2-\frac{x^6}6+\frac{x^{10}}{120}-\frac{x^{14}}{5040}+O(x^{16})\Bigr)\\ &=x^5-\frac{x^9}6+\frac{x^{13}}{120}-\frac{x^{17}}{5040}+O(x^{19}) \end{align}

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