What is an inverse function?

The standard formula for the derivative of an inverse function, which follows easily from the Chain Rule, is: if $f(x) = y,$ then $(f^{-1})'(y)f'(x) = 1.$ Note that $f(0) = 2.$ So $\frac1{(f^{-1})'(2)} = f'(0).$

Now $f(x) = 2e^x + e^x F(x),$ where $F(x) = \int_0^x \sqrt{t^4+1} \, dt.$ Differentiating, we get $\begin{aligned} f'(x) &= 2e^x + e^x F'(x) + e^x F(x) \\ f'(0) &= 2 + F'(0) + F(0) \end{aligned}$ Now $F(0) = 0,$ and by the Fundamental Theorem of Calculus , $F'(0) = \sqrt{0^4+1} = 1.$ So the answer is $2+1+0 = \fbox{3}.$

$f(0)=2$ is immediate from the problem. Therefore ${ f }^{ -1 }(2)=0$ follows. Now, differentiating the original equation, ${ e }^{ -x }f'(x)-{ e }^{ -x }f(x)=\sqrt { { x }^{ 4 }+1 }$ . Put ${ f }^{ -1 }(x)$ instead of $x$ . Resulting equation would be ${ e }^{ -{ f }^{ -1 }(x) }f'({ f }^{ -1 }(x))-{ e }^{ -{ f }^{ -1 }(x) }f({ f }^{ -1 }(x))=\sqrt { { ({ f }^{ -1 }(x)) }^{ 4 }+1 }\longrightarrow 1$ . Since ${ f }^{ -1 }$ is the inverse of $f$ , $f({ f }^{ -1 }(x))=x$ . Differentiating, $f'({ f }^{ -1 }(x))*(({ f }^{ -1 })'(x))=1$ . Therefore, $f'({ f }^{ -1 }(x))=\frac { 1 }{ ({ f }^{ -1 })'(x) }$ . Simplifying the equation 1, $\frac { { e }^{ -{ f }^{ -1 }(x) } }{ ({ f }^{ -1 })'(x) } -{ e }^{ -{ f }^{ -1 }(x) }x=\sqrt { { ({ f }^{ -1 }(x)) }^{ 4 }+1 }$ . Plugging in 2, we will get $\frac { 1 }{ ({ f }^{ -1 })'(x) } =3$

In one word we need to only f'(0)....as (f^-1(f(x)))' = 1/f'(x). So diffentiating with leibnitz rule and using that f(0)=2. We see that (e^-x)f'(x) - (e^-x)f(x) = (1+x^4)^1/2. Putting x=0 we have f'(0) - 2 = 1. So f'(0) = 3