Find the equation of tangent to the `y = F(x)` at `x = 1`, where `F(x) = int_(x)^(x^(3))(dt)/(sqrt(1+t^(2)))`

Answer

To find the equation of the tangent to the curve \( y = F(x) \) at \( x = 1 \), where

\[ F(x) = \int_{x}^{x^3} \frac{dt}{\sqrt{1+t^2}}, \]

we will follow these steps:

### Step 1: Calculate \( F(1) \)

Substituting \( x = 1 \):

\[ F(1) = \int_{1}^{1^3} \frac{dt}{\sqrt{1+t^2}} = \int_{1}^{1} \frac{dt}{\sqrt{1+t^2}}. \]

Since the upper and lower limits of the integral are the same, we have:

\[ F(1) = 0. \]

### Step 2: Find the derivative \( F'(x) \)

To find the derivative \( F'(x) \), we use the Leibniz rule for differentiation under the integral sign:

\[ F'(x) = \frac{d}{dx} \left( \int_{x}^{x^3} \frac{dt}{\sqrt{1+t^2}} \right) = \frac{d}{dx} \left( \int_{x^3} \frac{dt}{\sqrt{1+t^2}} - \int_{x} \frac{dt}{\sqrt{1+t^2}} \right). \]

Using the Fundamental Theorem of Calculus:

\[ F'(x) = \frac{d}{dx} \left( \int_{x^3} \frac{dt}{\sqrt{1+t^2}} \right) - \frac{d}{dx} \left( \int_{x} \frac{dt}{\sqrt{1+t^2}} \right). \]

This gives:

\[ F'(x) = \frac{d}{dx} \left( \frac{x^3}{\sqrt{1+(x^3)^2}} \cdot \frac{d}{dx}(x^3) \right) - \frac{d}{dx} \left( \frac{x}{\sqrt{1+x^2}} \cdot \frac{d}{dx}(x) \right). \]

Calculating these derivatives, we have:

\[ F'(x) = \frac{3x^2}{\sqrt{1+x^6}} - \frac{1}{\sqrt{1+x^2}}. \]

### Step 3: Evaluate \( F'(1) \)

Now, substituting \( x = 1 \):

\[ F'(1) = \frac{3(1^2)}{\sqrt{1+1^6}} - \frac{1}{\sqrt{1+1^2}} = \frac{3}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{3 - 1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}. \]

### Step 4: Write the equation of the tangent line

The equation of the tangent line at the point \( (x_1, y_1) = (1, 0) \) with slope \( m = F'(1) = \sqrt{2} \) is given by:

\[ y - y_1 = m(x - x_1). \]

Substituting the values:

\[ y - 0 = \sqrt{2}(x - 1). \]

This simplifies to:

\[ y = \sqrt{2}x - \sqrt{2}. \]

### Final Equation of the Tangent

Thus, the equation of the tangent to the curve \( y = F(x) \) at \( x = 1 \) is:

\[ y = \sqrt{2}x - \sqrt{2}. \]

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