Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ajmartinez Group Title

Can someone walk me through calculating the dericative with respect to x of arctan (x/sqrt(1-x^2) please?

  • 6 months ago
  • 6 months ago

  • This Question is Closed
  1. chrisfso Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, let me try... \[y=\tan ^{-1}\frac{ x }{ \sqrt{1-x^{2}} }\] take tan of each side, rewrite fraction as a product with negative exponent \[\tan y=x(1-x ^{2})^{\frac{ -1 }{ 2}}\] implicit differentiation, use product rule on right side derivative of tan x is \[\sec ^{2}x\] \[\sec ^{2}y y \prime = (1) (1-x ^{2})^{\frac{ -1 }{ 2 }}+x(\frac{ -1 }{ 2 })(1-x ^{2})^{\frac{ -3 }{ 2}}(2x)\] \[\sec ^{2}y y \prime=(1-x ^{2})^{\frac{ -1 }{ 2 }}-x ^{2}(1-x ^{2})^{\frac{ -3 }{ 2 }}\] \[y \prime = \frac{ (1-x ^{2})^{\frac{ -1 }{ 2 }}-x ^{2}(1-x ^{2})^{\frac{ -3 }{ 2 }} }{\sec ^{2}y}\] we know that cos = 1/sec and 1/cos = sec \[ y \prime=((1-x ^{2})^{\frac{ -1 }{ 2 }}-x ^{2}(1-x ^{2})^{\frac{ -3 }{ 2 }})\cos ^{2}y\] ok, all we need is y in terms of x we know from the beginning that \[\tan y=x(1-x ^{2})^{\frac{ -1 }{ 2}}\] so build a right triangle with adjacent side 1, opp side \[x(1-x ^{2})^{\frac{ -1 }{ 2}}\] solve for hypotenuse = \[hypotenuse=\sqrt{1^{2}+(x(1-x ^{2})^\frac{ 1 }{ 2 })^{2}}\] \[hypotenuse=\sqrt{1+x ^{2}(1-x ^{2})}\] therefore we can now solve for cos = adjacent over hypotenuse \[\cos y = \frac{ 1 }{ \sqrt{1+x ^{2}(1-x ^{2})} }\] \[\cos ^{2} y = \frac{ 1 }{ 1+x ^{2}(1-x ^{2})}\] plugging back into the y prime equation \[y \prime=\frac{ ((1-x ^{2})^{\frac{ -1 }{ 2 }}-x ^{2}(1-x ^{2})^{\frac{ -3 }{ 2 }}) }{ 1+x ^{2}(1-x ^{2}) }\] any experts to tell us if this is correct?

    • 6 months ago
  2. davidstuartbruce Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I'd do it like this: \[\tan y = x/\sqrt{1-x ^{2}}\] Now make a right triangle based on this relationship: |dw:1398448709675:dw| So that means \[\sin y = x\] Differentiating, \[\cos yy' = 1\] or\[y' = 1/\cos y\] Since \[\sin ^{2}\Theta + \cos ^{2}\theta = 1\] we have: \[y' = 1/\sqrt{1-\sin ^{2}y}\] And above we had sin y = x, so: \[y' = 1/\sqrt{1-x ^{2}}\]

    • 6 months ago
  3. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    fyi @chrisfso you dropped a minus sign \[ \sec ^{2}y\ y \prime = (1) (1-x ^{2})^{\frac{ -1 }{ 2 }}+x(\frac{ -1 }{ 2 })(1-x ^{2})^{\frac{ -3 }{ 2}}(-2x) \] which makes the next line \[ \sec ^{2}y\ y \prime=(1-x ^{2})^{\frac{ -1 }{ 2 }}+x ^{2}(1-x ^{2})^{\frac{ -3 }{ 2 }} \] if you use cos^2 y = (1-x^2), and multiply both sides by cos^2 y , you get \[ y' = (1-x ^{2})^{\frac{ 1 }{ 2 }}+x ^{2}(1-x ^{2})^{\frac{ -1 }{ 2 }} \] if you simplify , you get david's result

    • 6 months ago
  4. tderifield Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Without taking anything away from the great explanations above, if you want to see one that explicates the solution given in the handout, please check out the attached explanation from creeksider provided in a different thread. It fills in the missing steps in the answer. Thanks again, creeksider!

    • 6 months ago
  5. chrisfso Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks phi!

    • 5 months ago
  6. arnavguddu Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    write z=x/sqrt(1-x^2) so f = arctan z f' = z'/(1+z^2) = z'/(x^2/(1-x^2)) = (1-x^2)z'/x^2 write q = sqrt(1-x^2) so z = x/q z' = 1/q - x q'/q^2 = 1/sqrt(1-x^2) - x q'/(1-x^2) write r = 1- x^2 then q = sqrt(r) q' = r'/sqrt(r) = r'/sqrt(1-x^2) r= 1-x^2 r' = 0 - 2x = -2x this is a easy to understand way.....if u get it...u can do the substitution in the head without writing.....and continue..

    • 5 months ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.