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One of the practice problems in my Calculus book is as follows: The graph of y=$8/(x^2-4)$ is called the Witch of Agnesi.

(a) Find y'

d/dx (u/v) = (v du/dx - u dv/dx)/($v^2$)
$=((x^2+4)(0)-(8)(2x))/((x^2+4)^2)$
$=(-16x)/((x^2+4)^2)$

Assuming I've gotten (a) correct, I then am asked (b) Find the equation of the tangent (in slope-intercept form) at the point (2,1).

Can anyone help me get rolling on this one?

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  • $\begingroup$ one thing thats missing is the derivation of the equation $d/dx (u/v) = (v du/dx - u dv/dx)/v^2$. $\endgroup$ – John Joy Oct 13 '15 at 23:22
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Undefined. Using your equation given and plugging $x=2$ gives $\frac80$ or undefined. The proper equation for the witch of agnesi is $\frac8{x^2+4}$.

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Here's a method to get you rolling:

The derivative of $f(x)$ is the slope of the tangent line at the point $(x, f(x))$. You already found the derivative to be $-\frac{16x}{(x^2+4)^2}$, so this is the slope of the tangent line at any point $(x, f(x))$.

To find the equation of the tangent line at $(2, 1)$, you first need to find the slope of the tangent line at $(2, 1)$, let's call it $m$. After you find $m$, you then plug $m$ and the point $(2, 1)$ into the point-slope formula and solve for $y$ $$ y - y_1 = m(x - x_1).$$

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  • $\begingroup$ Ok, but how do I find $m$ to proceed? $\endgroup$ – Analytic Lunatic Oct 13 '15 at 20:41
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    $\begingroup$ @AnalyticLunatic To find $m$, you have to evaluate the derivative when $x$ is $2$ (plug $2$ in for $x$). This will give you the slope of the tangent line at $x=2$. $\endgroup$ – josh Oct 13 '15 at 22:50
  • $\begingroup$ @AnalyticLunatic Feel free to accept my answer as the final answer by clicking the check mark under next to the voting arrows. Thanks! $\endgroup$ – josh Oct 27 '16 at 14:35
  • $\begingroup$ I do not know how you picked it.[correct equation][1] y= f(x) is in Mathworld [1]: mathworld.wolfram.com/WitchofAgnesi.html Negative sign in denominator is incorrect. You may be barking the wrong tree, but no harm, procedure is clear from others' responses. $\endgroup$ – Narasimham Feb 18 '17 at 12:11

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