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ajmartinez

  • 8 months ago

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

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  1. chrisfso
    • 7 months ago
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    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?

  2. davidstuartbruce
    • 7 months ago
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    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: \[y' = 1/\sqrt{1-x ^{2}}\]

  3. phi
    • 7 months ago
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    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

  4. tderifield
    • 7 months ago
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    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!

  5. chrisfso
    • 7 months ago
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    Thanks phi!

  6. arnavguddu
    • 7 months ago
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    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..

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