Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

OCW Scholar - Single Variable Calculus
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
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?
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: \[y' = 1/\sqrt{1-x ^{2}}\]
  • phi
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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!
Thanks phi!
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..

Not the answer you are looking for?

Search for more explanations.

Ask your own question