# Taylor series expansion of $\sin(2x^2)$

I am doing homework an online calc class and having a tough time getting hold of the prof, so I was hoping someone here could let me know what I am doing wrong. I have to find the first few terms for a Taylor series expansion of the function $$f(x)=\sin(2x^2)$$

I'm given that $f(x)=\sin(x)$ has the expansion $$x-x^3/3!+x^5/5!$$ So I decide to sub $2x^2$ for $x$ in each term and get $$2x^2-2x^6/6+2x^{10}/120$$ which is $$2x^2-x^6/3+x^{10}/60$$. but when I use the math software provided by the class to check my answer (as requested in assignment). I get $$2x^2-(4/3)x^6 + \cdots$$ and I can't figure out why the term has a $4/3$ coefficient instead of a $1/3$ coefficient above. Can someone help me with what I am doing wrong here ?

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Thanks a bunch. I knew it was something easy. Also thanks for the editing that provided actual exponents. Is there a help file for how to do that ? – chrisfs Mar 29 '12 at 22:59

$$\frac{(2x^2)^3}{6} = \frac{8x^6}{6} = \frac{4x^6}{3}$$
You forgot to cube the $2$.