Finding a higher-degree Maclaurin polynomial from $f(0)=1$ and $f'(x)=1+(f(x))^{10}$

Question

Let $f(x)$ be a function such that $f(0)=1$ and there exists a neighborhood of $x=0$ in which $f'(x)=1+(f(x))^{10}$. What is the 3rd-degree Maclaurin polynomial of $f(x)$?

Answer 1

First, you need to know how to apply the chain rule.

$f''(x)=\frac {d}{dx}\left (1+\left (f(x) \right)^{10} \right)=0 + 10 \left (f(x) \right)^{9} f'(x)= 10 f'(x) \left (f(x) \right)^{9}$

To find $f'''(x)$ you can do one of these:

• substitute $f'(x)=1+\left (f(x) \right)^{10}$ into the expression above to get an expression for $f''(x)$ purely in terms of $f(x)$ and then differentiate that.

• use the product rule on the expression above (which I think is easier).