Question

Find the tangent to the witch of Agnesi
y=8/(4+x^2) at the point (2,1).

Answer 1

So you want to find the line tangent to the function y = 8/(x^2+4) at point x = 2. So, in other words, the function of a line which has slope equal to the value of y' at x = 2. And that intersects point (2,1). Do you know how to start the question?

Answer 2

I think I'm supposed to start by finding the slope by differentiating. However, it seems I am doing that part wrong since I can't get the right answer.

Answer 3

What did you get for y'?

Answer 4

-16/((x^2+4)^2)

Answer 5

Close.\[\frac{ d }{ dx }[\frac{ 8 }{ 4+x^2 }] = 8 \frac{ d }{ dx }[\frac{ 1 }{ 4+x^2 }] = 8 * -2 \frac{ x }{( x^2+4)^2 } = \frac{ -16x }{ (x^2+4)^2 }\]

Answer 6

Is this the chain rule?

Answer 7

Yes, but you could also use the quotient rule if you wanted to.

Answer 8

for quotient rule you would have f(x) = 8; f'(x) = 0; g(x) = x^2 + 4; g'(x) = 2x so \[\frac{ dy }{ dx } = \frac{ 0(x^2+4) - 8*2x }{ (x^2+4)^2 } = \frac{ -16x }{ (x^2+4)^2 }\]

Answer 9

I see what I was do ng wrong now. Thanks for helping me. :)

Answer 10

No problem :). Do you know what to do from there?

Answer 11

Yes. I just finished the rest and got the right answer.