Question

Find the d/dx and tangent

Answer 1

\[\LARGE \frac{|x|}{\sqrt{5-x^2}}\]
at (2, 2)

Answer 2

this problem looks harder than it is. The function \[f(x)=\frac{ |x| }{ \sqrt{5-x^2} }\] simplifies to \[f(x)=\frac{ x }{ \sqrt{5-x^2} }\]

Answer 3

for x > 0.

Apply the Quotient Rule to find the derivative of f(x).

Then substitute x=2 and you'll have your answer (that is, the slope of the TL at (2,2).

continue onward to find the equation of the tangent line to the curve at tht point.

Answer 4

An absolute value always defines a non-differentiable point, so you'll have two derivatives.

Answer 5

True. But only the derivative of the right-hand half of the function (that is, the half defined only for positive x-values) is needed in this problem. Can you explain why that is?

Answer 6

Divide by 0.

Answer 7

Yup, because that non-differentiable point, if we look at the definition of abs. value will be in the x=0.
The definition of abs. value is:
\[\left| \alpha \right|=(\alpha) , x>0\]
\[\left| \alpha \right|=-(\alpha), x<0 \]
So, doing that to the function and since the denominator is a function, and by definition:
\[1)\left| x \right|=(x),x>0\]
\[2)\left| x \right|=-(x),x<0\]
then:
\[1)\frac{ \left| x \right| }{ \sqrt{5-x ^{2}} }=\frac{ (x) }{ \sqrt{5-x ^{2}} }, x>0\]
\[2)\frac{ \left| x \right| }{ \sqrt{5-x ^{2}} }=\frac{ -(x) }{ \sqrt{5-x ^{2}} }\]
if you derivate 1 and 2 individually you'll find 2 different derivatices, meaning that x=0 is a non differentiable point.

Answer 8

Actually, it's because we need the derivative ONLY for positive x, since the point of tangency, (2,2) has x-coordinate +2.

@luigi0210 : Mind calculating the appropriate derivative? Use the Quotient Rule. when you're done with that, let x=2. OK?

Answer 9

forgot the x<0 there.

Answer 10

Would you mind explaining why you believe we need the derivative for negative x, given that the point of tangency is at x=+2, y=+2?

@Luigi0210 : the ball's in your court; hope to hear from you soon.

Answer 11

actually we don't, but it's a step to take, you can't simply say "Oh, it's that" without stating it.
When you find both derivatives and the non-differentiable point you take the the corresponding derivative of the point in question.
I hate being unconsistent or skipping steps, that makes any exercise look dull and incomplete.

Answer 12

@luigi0210 : I sincerely hope that some of these postings will be of good use to you. I won't be posting any more replies until after we've heard back from you. Best wishes. MM

Answer 13

I've tried it, still got the wrong answer ._.
I think I differentiated it wrong..

Answer 14

Best I can suggest at this point is that you find a way to share your work with your helpers, e. g., to take a cell phone photo of it and to upload that image file to OpenStudy. Have you done that before? Unfortunately, I have to get off the 'Net now, but would be happy to continue working with you tomorrow on this problem.

Answer 15

f'(x) = sqrt(x^2) / (x * (5-x^2)^(3/2) ) will take care of both cases

Answer 16

I'll just set up the derivative; if you like, you Luigi0210 and sourwing, could compare our three results:\[\frac{ d }{ dx }\frac{ x }{ (5-x ^{2})^.5 }\]

Answer 17

\[= \frac{ (5-x ^{2})^{.5}(1)-(x)*.5(5-x ^{2} )^{-.5}(-2x)}{ 5-x ^{2} }\]

Answer 18

Here I've used the quotient, power and chain rules.

Answer 19

And actually I figured it out last night when you left. Thanks for the help :)