A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 2 years ago
Can someone walk me through calculating the dericative with respect to x of arctan (x/sqrt(1x^2) please?
anonymous
 2 years ago
Can someone walk me through calculating the dericative with respect to x of arctan (x/sqrt(1x^2) please?

This Question is Closed

chrisfso
 2 years ago
Best ResponseYou've already chosen the best response.0ok, let me try... \[y=\tan ^{1}\frac{ x }{ \sqrt{1x^{2}} }\] take tan of each side, rewrite fraction as a product with negative exponent \[\tan y=x(1x ^{2})^{\frac{ 1 }{ 2}}\] implicit differentiation, use product rule on right side derivative of tan x is \[\sec ^{2}x\] \[\sec ^{2}y y \prime = (1) (1x ^{2})^{\frac{ 1 }{ 2 }}+x(\frac{ 1 }{ 2 })(1x ^{2})^{\frac{ 3 }{ 2}}(2x)\] \[\sec ^{2}y y \prime=(1x ^{2})^{\frac{ 1 }{ 2 }}x ^{2}(1x ^{2})^{\frac{ 3 }{ 2 }}\] \[y \prime = \frac{ (1x ^{2})^{\frac{ 1 }{ 2 }}x ^{2}(1x ^{2})^{\frac{ 3 }{ 2 }} }{\sec ^{2}y}\] we know that cos = 1/sec and 1/cos = sec \[ y \prime=((1x ^{2})^{\frac{ 1 }{ 2 }}x ^{2}(1x ^{2})^{\frac{ 3 }{ 2 }})\cos ^{2}y\] ok, all we need is y in terms of x we know from the beginning that \[\tan y=x(1x ^{2})^{\frac{ 1 }{ 2}}\] so build a right triangle with adjacent side 1, opp side \[x(1x ^{2})^{\frac{ 1 }{ 2}}\] solve for hypotenuse = \[hypotenuse=\sqrt{1^{2}+(x(1x ^{2})^\frac{ 1 }{ 2 })^{2}}\] \[hypotenuse=\sqrt{1+x ^{2}(1x ^{2})}\] therefore we can now solve for cos = adjacent over hypotenuse \[\cos y = \frac{ 1 }{ \sqrt{1+x ^{2}(1x ^{2})} }\] \[\cos ^{2} y = \frac{ 1 }{ 1+x ^{2}(1x ^{2})}\] plugging back into the y prime equation \[y \prime=\frac{ ((1x ^{2})^{\frac{ 1 }{ 2 }}x ^{2}(1x ^{2})^{\frac{ 3 }{ 2 }}) }{ 1+x ^{2}(1x ^{2}) }\] any experts to tell us if this is correct?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I'd do it like this: \[\tan y = x/\sqrt{1x ^{2}}\] 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: \[y' = 1/\sqrt{1x ^{2}}\]

phi
 2 years ago
Best ResponseYou've already chosen the best response.0fyi @chrisfso you dropped a minus sign \[ \sec ^{2}y\ y \prime = (1) (1x ^{2})^{\frac{ 1 }{ 2 }}+x(\frac{ 1 }{ 2 })(1x ^{2})^{\frac{ 3 }{ 2}}(2x) \] which makes the next line \[ \sec ^{2}y\ y \prime=(1x ^{2})^{\frac{ 1 }{ 2 }}+x ^{2}(1x ^{2})^{\frac{ 3 }{ 2 }} \] if you use cos^2 y = (1x^2), and multiply both sides by cos^2 y , you get \[ y' = (1x ^{2})^{\frac{ 1 }{ 2 }}+x ^{2}(1x ^{2})^{\frac{ 1 }{ 2 }} \] if you simplify , you get david's result

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Without 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!

arnavguddu
 2 years ago
Best ResponseYou've already chosen the best response.0write z=x/sqrt(1x^2) so f = arctan z f' = z'/(1+z^2) = z'/(x^2/(1x^2)) = (1x^2)z'/x^2 write q = sqrt(1x^2) so z = x/q z' = 1/q  x q'/q^2 = 1/sqrt(1x^2)  x q'/(1x^2) write r = 1 x^2 then q = sqrt(r) q' = r'/sqrt(r) = r'/sqrt(1x^2) r= 1x^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..
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.