Find the Maclaurin series for f(x)= x^2/sqrt(x^5+1)

Question

I really have no idea how to go about this, could really use some help, thank you!

Answer 1

f(x) = x²/(x⁵+1)^1/2 equal to 0 at x=0.

f'(x)=(−0.5x⁶+2x)/(x⁵+1)^3/2 equal to 0 at x=0.

f''(x)=(0.75x¹⁰−16x⁵+2)/(x⁵+1)^5/2 equal to 2 at x=0.

f'''(x)=(−1.875x¹⁴+127.5x⁹−105x⁴)/(x⁵+1)^7/2 equal to 0 at x=0.

f^IV(x)=(6.5625x¹⁸−1110x¹³+2565x⁸−420x³)/(x⁵+1)^9/2 equal to 0 at x=0.

The Maclaurin Series Generated By f(x) is given by: f(0)+(f'(0)/1!)x¹+(f''(0)/2!)x²+(f'''(0)/3!)x³+(f^IV(0)/4!)x⁴

+...+(fⁿ(0)/n!)xⁿ+...

For f(x) = x²/(x⁵+1)^1/2, write f(x)=0+(0/1!)x¹+(2/2!)x²+(0/3!)x³+(0/4!)x⁴+...

Assuming that every term after (0/4!)x⁴ comes to 0, obtain f(x) [or x²/(x⁵+1)^1/2] =(2/2!)x² or x² which is true at x=0.