8. [-/1.07 Points] DETAILS MY NOTES SCALCET9 3.4.062. A graphing calculator is recommended. The curve $$ y=\frac{|x|}{\sqrt{2-x^{2}}} $$ is called a bullet-nose curve. (a) Find an equation of the tangent line to this curve at the point $$ (1,1) $$. (a)

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**Step 1. Express the function for $$ x > 0 $$**

For $$ x > 0 $$, the absolute value simplifies as $$ |x| = x $$. Thus, the function becomes

$$
y=\frac{x}{\sqrt{2-x^2}}.
$$

**Step 2. Differentiate the function using the product rule**

Rewrite the function as

$$
y = x \cdot (2-x^2)^{-1/2}.
$$

Let
$$
u(x) = x \quad \text{and} \quad v(x) = (2-x^2)^{-1/2}.
$$

Then, by the product rule,

$$
y' = u'(x) \cdot v(x) + u(x) \cdot v'(x).
$$

We have
$$
u'(x) = 1.
$$

**Step 3. Differentiate $$ v(x) $$ using the chain rule**

Given

$$
v(x) = (2-x^2)^{-1/2},
$$

let 
$$
w(x)=2-x^2.
$$
Then, $$ v(x) = w(x)^{-1/2} $$ and by the chain rule:

$$
v'(x) = -\frac{1}{2} w(x)^{-3/2} \cdot w'(x).
$$

Differentiate $$ w(x) $$:

$$
w'(x) = -2x.
$$

Substitute back:

$$
v'(x) = -\frac{1}{2}(2-x^2)^{-3/2} \cdot (-2x) = x (2-x^2)^{-3/2}.
$$

**Step 4. Combine the derivatives**

Substitute $$ u'(x) $$, $$ v(x) $$, and $$ v'(x) $$ into the product rule:

$$
y' = 1\cdot (2-x^2)^{-1/2} + x \cdot \left[ x(2-x^2)^{-3/2} \right].
$$

Simplify:

$$
y' = (2-x^2)^{-1/2} + x^2 (2-x^2)^{-3/2}.
$$

Factor out $$ (2-x^2)^{-3/2} $$:

$$
y' = (2-x^2)^{-3/2} \left[(2-x^2) + x^2\right] = (2-x^2)^{-3/2} \cdot 2.
$$

Thus,

$$
y' = \frac{2}{(2-x^2)^{3/2}}.
$$

**Step 5. Evaluate the derivative at $$ x=1 $$**

Substitute $$ x = 1 $$ into the derivative:

$$
y'(1) = \frac{2}{(2-1^2)^{3/2}} = \frac{2}{(2-1)^{3/2}} = \frac{2}{1^{3/2}} = 2.
$$

So, the slope of the tangent line at $$ (1,1) $$ is $$ m = 2 $$.

**Step 6. Find the equation of the tangent line**

The tangent line at $$ (x_0, y_0) = (1,1) $$ with slope $$ m = 2 $$ is given by the point-slope form:

$$
y - y_0 = m(x - x_0).
$$

Substitute the values:

$$
y - 1 = 2(x - 1).
$$

Solve for $$ y $$:

$$
y = 2x - 2 + 1,
$$
$$
y = 2x - 1.
$$

Thus, the equation of the tangent line is

$$
y = 2x - 1.
$$
The equation of the tangent line to the curve $$ y = \frac{|x|}{\sqrt{2 - x^2}} $$ at the point $$ (1,1) $$ is $$ y = 2x - 1 $$.

[-/1.07 Points] DETAILS MY NOTES SCALCET9 3.4.062.
A graphing calculator is recommended.
The curve is called a bullet-nose curve.
(a) Find an equation of the tangent line to this curve at the point .
(a)

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[-/1.07 Points] DETAILS MY NOTES SCALCET9 3.4.062. A graphing calculator is recommended. The curve is called a bullet-nose curve. (a) Find an equation of the tangent line to this curve at the point . (a)