We are given the function:

$g(x) = \int_0^{1/x} \frac{t}{\sqrt{1 + t^4}} \, dt$

and we need to find its derivative $g'(x)$.

Step 1: Apply the Fundamental Theorem of Calculus (FTC) Part I

The Fundamental Theorem of Calculus states that if:

$G(x) = \int_{a}^{h(x)} f(t) \, dt$

then:

$G'(x) = f(h(x)) \cdot h'(x).$

In our case:

$g(x) = \int_0^{1/x} \frac{t}{\sqrt{1 + t^4}} \, dt$

So, using the FTC:

$g'(x) = f(1/x) \cdot \frac{d}{dx} \left( \frac{1}{x} \right)$

Step 2: Evaluate $f(1/x)$

We define the function inside the integral as:

$f(t) = \frac{t}{\sqrt{1 + t^4}}$

Substituting $t = 1/x$:

$f(1/x) = \frac{(1/x)}{\sqrt{1 + (1/x)^4}}$

$= \frac{1/x}{\sqrt{1 + 1/x^4}}.$

Step 3: Compute $\frac{d}{dx} (1/x)$

We know that:

$\frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2}.$

Step 4: Multiply the Terms

$g'(x) = \left( \frac{1/x}{\sqrt{1 + 1/x^4}} \right) \cdot \left( -\frac{1}{x^2} \right).$

$= -\frac{(1/x)}{x^2 \sqrt{1 + 1/x^4}}.$

$= -\frac{1}{x^3 \sqrt{1 + 1/x^4}}.$

Step 5: Simplify Further

Rewriting the denominator:

$1 + \frac{1}{x^4} = \frac{x^4 + 1}{x^4}.$

Taking the square root:

$\sqrt{1 + 1/x^4} = \sqrt{\frac{x^4 + 1}{x^4}} = \frac{\sqrt{x^4 + 1}}{x^2}.$

Substituting back:

$g'(x) = -\frac{1}{x^3} \cdot \frac{x^2}{\sqrt{x^4 + 1}}.$

$= -\frac{x^2}{x^3 \sqrt{x^4 + 1}}.$

$= -\frac{1}{x \sqrt{x^4 + 1}}.$

Final Answer:

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

Let me know if you want more details or have any questions!

Related Questions:

1. How does the Fundamental Theorem of Calculus help in differentiation?
2. What happens if the upper limit was $x^2$ instead of $1/x$?
3. How do we differentiate an integral with both upper and lower limits as functions?
4. What are some real-world applications of the Fundamental Theorem of Calculus?
5. How do we approach similar problems using Leibniz’s rule for differentiation under the integral sign?

Tip:

When differentiating an integral with a variable upper bound, always apply FTC carefully and remember to multiply by the derivative of the upper bound.