Could you please give me some hint how to find 3-th degree Maclaurin polynomial of f(x) given f(0)=1 and for all $0<x<\lambda$ $f'(x)=1+f(x)^{10}$.
If $\lim_{x\to0}f(x)=f(0)=1$ then $\lim_{x\to 0}f'(x)=f'(0)=2$.
But how to calculate second and third derivative of f(x) ?
Thanks.
To calculate the second and third derivatives use the chain and product rule.
We have $$ f''(x) = [1+f(x)^{10}]' = 10 f(x)^9 f'(x) $$ and $$ f'''(x) = [10 f(x)^9 f'(x)]' = 90 f(x)^8 [f'(x)]^2 + 10 f(x)^9 f''(x) $$ You should be able to plug in from here.