The equation of the tangent to the curve `y= int_(x^4)^(x^6) (dt)/( sqrt( 1+t^2) )` at `x=1` is

Options

A. `sqrt(2) y+ 1 =x`
B. `sqrt(3) x+1 =y`
C. `sqrt(3) x+1 + sqrt3`
D. none of these

Answer

To find the equation of the tangent to the curve \( y = \int_{x^4}^{x^6} \frac{dt}{\sqrt{1+t^2}} \) at \( x = 1 \), we will follow these steps:

### Step 1: Differentiate the integral using Leibniz's rule
We need to differentiate the function \( y \) with respect to \( x \) using Leibniz's rule. According to Leibniz's rule, if we have an integral with variable limits, the derivative is given by:

\[ \frac{dy}{dx} = \frac{d}{dx}\left( \int_{a(x)}^{b(x)} f(t) dt \right) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \]

In our case, \( a(x) = x^4 \), \( b(x) = x^6 \), and \( f(t) = \frac{1}{\sqrt{1+t^2}} \).

### Step 2: Apply Leibniz's rule
Now we apply the rule:

\[ \frac{dy}{dx} = \frac{1}{\sqrt{1+(x^6)^2}} \cdot \frac{d}{dx}(x^6) - \frac{1}{\sqrt{1+(x^4)^2}} \cdot \frac{d}{dx}(x^4) \]

Calculating the derivatives:

\[ \frac{d}{dx}(x^6) = 6x^5 \quad \text{and} \quad \frac{d}{dx}(x^4) = 4x^3 \]

Substituting these into the equation gives:

\[ \frac{dy}{dx} = \frac{6x^5}{\sqrt{1+x^{12}}} - \frac{4x^3}{\sqrt{1+x^8}} \]

### Step 3: Evaluate at \( x = 1 \)
Now we need to evaluate \( \frac{dy}{dx} \) at \( x = 1 \):

\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{6 \cdot 1^5}{\sqrt{1+1^{12}}} - \frac{4 \cdot 1^3}{\sqrt{1+1^8}} = \frac{6}{\sqrt{2}} - \frac{4}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 4: Find the value of \( y \) at \( x = 1 \)
Next, we find the value of \( y \) at \( x = 1 \):

\[ y = \int_{1^4}^{1^6} \frac{dt}{\sqrt{1+t^2}} = \int_{1}^{1} \frac{dt}{\sqrt{1+t^2}} = 0 \]

### Step 5: Write the equation of the tangent line
We have the slope \( m = \sqrt{2} \) and the point \( (1, 0) \). The equation of the tangent line can be written using the point-slope form:

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

Substituting \( (x_1, y_1) = (1, 0) \) and \( m = \sqrt{2} \):

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

This simplifies to:

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

### Final Answer
The equation of the tangent to the curve at \( x = 1 \) is:

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