Let’s solve the problem step by step.

The given function is:

$g(x) = \int_2^x \frac{1}{\sqrt{1 + t^4}} \, dt$

Part (a) Find $g'(x)$.

To differentiate $g(x)$, we use the Fundamental Theorem of Calculus, which states that if: $h(x) = \int_a^x f(t) \, dt$ then: $h'(x) = f(x).$

Applying this to $g(x)$, we get: $g'(x) = \frac{1}{\sqrt{1 + x^4}}.$

Part (b) Find $(g^{-1})'(0)$.

To solve for $(g^{-1})'(0)$, we use the Inverse Function Theorem, which states that if $y = g(x)$ and $g'(x) \neq 0$, then: $(g^{-1})'(y) = \frac{1}{g'(g^{-1}(y))}.$

Since we are asked for $(g^{-1})'(0)$, we need to determine the value of $g^{-1}(0)$. Let $g(a) = 0$, so $g^{-1}(0) = a$.

From the definition of $g(x)$: $g(x) = \int_2^x \frac{1}{\sqrt{1 + t^4}} \, dt.$

For $g(x) = 0$, it must be that $x = 2$, since the integral from 2 to itself is zero. Thus: $g^{-1}(0) = 2.$

Next, we compute $(g^{-1})'(0)$ using the formula: $(g^{-1})'(0) = \frac{1}{g'(g^{-1}(0))}.$

Substituting $g^{-1}(0) = 2$: $(g^{-1})'(0) = \frac{1}{g'(2)}.$

From Part (a), $g'(x) = \frac{1}{\sqrt{1 + x^4}}$. Substituting $x = 2$: $g'(2) = \frac{1}{\sqrt{1 + 2^4}} = \frac{1}{\sqrt{1 + 16}} = \frac{1}{\sqrt{17}}.$

Thus: $(g^{-1})'(0) = \frac{1}{\frac{1}{\sqrt{17}}} = \sqrt{17}.$

Final Answers:

(a) $g'(x) = \frac{1}{\sqrt{1 + x^4}}$.

(b) $(g^{-1})'(0) = \sqrt{17}$.

Would you like a more detailed explanation of any step?

Related Questions:

1. What is the Fundamental Theorem of Calculus and how does it apply here?
2. How is the Inverse Function Theorem used to find derivatives of inverse functions?
3. How would you evaluate $g'(x)$ if the integral bounds were functions of $x$?
4. What is the interpretation of $g^{-1}(x)$ in terms of the graph of $g(x)$?
5. How would $g(x)$ change if the upper bound were a constant and the lower bound depended on $x$?

Tip:

When solving problems involving inverse functions, always ensure you properly evaluate the derivative of the original function at the correct point.