How do u do this ...is there an easy way to do it?

Question

find the coefficient of x^6 in the Taylor series expansion centered at the origin for the function f(x)= 3+7sinx^2

Answer 1

Note that the general form for the Taylor series is:

[n=0, ∞]∑(f⁽ⁿ⁾(0)/n! * x^n)

So all we have to do is find the 6th derivative of 3+7 sin (x²). Although, we don't really have to do that, since the taylor series for 3+7 sin (x²) may be obtained simply by substituting x² for x in the taylor series for 3+7 sin x. So the coefficient of the x^6 term in the series for 3+7 sin (x²) is the coefficient of the x³ term in the series for 3+ 7 sin x. That coefficient is:

(d³(3+7 sin x)/dx³)(0)/3!
-7 cos 0/3!
-7/6