\(f(0)=4\\ f'(0)=5\\ f''(0)=-1\\ f'''(0)=-\frac{15}{2}\\ f^{(4)}(0)=23\\ \text{Let g be a function such that}\;g(x)=f(x^3).\\ \text{Write the fifth-degree Taylor polynomial for g', the derivative of g, about x=0.}\)
So, I thought you had to find g'(x). I used chain rule and found g'(x) to equal 3x2f'(x3).
However, when doing that, I get something different from the answer.
Attempt:
\(f(x)\approx4+5x-\frac{x^2}{2}-\frac{15x^3}{12}+\frac{23x^4}{24}\\ g(x)=f(x^3)\approx4+5x^3-\frac{x^6}{2}-\frac{15x^9}{12}+\frac{23x^{12}}{24}\\ f'(x^3)\approx15x^2-3x^5\\ g'(x)=3x^2f'(x^3)\approx3x^2(15x^2-3x^5)=45x^4-9x^7\)
Where did I go wrong? Thanks!
