Question

a curve like $y = \frac{1}{1 + 1x^{2}}$ is called a witch of maria agnesi. find the equation of the tangent line to this curve at the point $(-1,\frac{1}{2})$.
$y=$

Explanation:

Step1: Differentiate the function

Use the quotient - rule. If $y=\frac{u}{v}$ where $u = 1$ and $v=1 + x^{2}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Since $u^\prime = 0$ and $v^\prime=2x$, we have $y^\prime=\frac{0\times(1 + x^{2})-1\times2x}{(1 + x^{2})^{2}}=-\frac{2x}{(1 + x^{2})^{2}}$.

Step2: Find the slope of the tangent line at $x=-1$

Substitute $x = - 1$ into $y^\prime$. $y^\prime|_{x=-1}=-\frac{2\times(-1)}{(1+(-1)^{2})^{2}}=\frac{2}{4}=\frac{1}{2}$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-1,\frac{1}{2})$ and $m=\frac{1}{2}$. So $y-\frac{1}{2}=\frac{1}{2}(x + 1)$.

Step4: Simplify the equation

Expand the right - hand side: $y-\frac{1}{2}=\frac{1}{2}x+\frac{1}{2}$. Then add $\frac{1}{2}$ to both sides to get $y=\frac{1}{2}x + 1$.

Answer:

$y=\frac{1}{2}x + 1$