Question

The curve y = x/(1 + x2) is called a serpentine. find an equation of the tangent line to this curve at the point (2, 0.40). (round the slope and y-intercept to two decimal places.) y =

Answer 1

One way of solving this is using calculus: derivatives.

We have $y = \frac{x}{1 + x^{2}}$. Let's take this as a function of x.

So, 

$f(x) = \frac{x}{1 + x^{2}}$.

Now, we find it's derivative: f'(x). To do this, we use the quotient rule. The derivative [ f'(x) ] basically gives us the slope of the tangent line at a point x.

$f'(x) = \frac{1*(1 + x^{2}) - (2x *x)}{ (1 + x^{2} )^{2} } = \frac{1 + x^{2} - 2 x^{2} }{ (1 + x^{2} )^{2} } = \frac{1 - x^{2} }{ (1 + x^{2} )^{2} }$

Now, we evaluate the derivative at the point x = 2 to find the slope of the tangent line at that point.

$f'(2) = \frac{1- 2^{2} }{ (1 + 2^{2} )^{2} } = \frac{1-4}{ (5 )^{2} } = \frac{-3}{25} = -0.12$
Now, we have the slope and the x and y co-ordinates; thus, we can use the point slope form to find the equation of the tangent line at the point (2,0.40).

$y - (0.4) = ( -0.12) *(x-2)$

$y -0.4 = -0.12x + 0.24$

$y = -0.12x + 0.28$

That's the equation.