If $f(x)=e^{g(x)}$ and $g(x)=\int\limits_2^x\frac{dt}{1+t^4}$ then f'(2) is equal to:

0 $\frac{1}{17}$ $\frac{3}{17}$ $\frac{2}{17}$

$\frac{1}{17}$

$g(x)=\int\limits_2^x\frac{dt}{1+t^4}⇒g(2)=\int\limits_2^2\frac{dt}{1+t^4}=0;g'(x)=\frac{1}{1+x^4}⇒g'(2)=\frac{1}{1+2^4}=\frac{1}{17}$ and $f(x)=e^{g(x)}⇒f'(x)=e^{g(x)}.g'(x)⇒f(2)e^0.\frac{1}{17}=\frac{1}{17}$