anonymous
  • anonymous
Find an equation of the tangent line to the bullet-nose curve y=|x|/sqrt(2−x^2) at the point (1,1) I think that square root is what is confusing me on this question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
Since x>0, then |x|=x
myininaya
  • myininaya
You need to use quotient rule + chain rule + product rule + constant rule
myininaya
  • myininaya
\[\sqrt{2-x^2}=(2-x^2)^\frac{1}{2}\] \[[[(2-x^2)]^\frac{1}{2}]'=\frac{1}{2}(2-x^2)^{\frac{1}{2}-1}(2-x^2)' \text{ note: I applied chain rule here}\]

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anonymous
  • anonymous
do i then need to apply the quotient rule using the absolute value of x?
amistre64
  • amistre64
\[y=|x|*(2−x^2)^{-1/2} \] \[D[y=|x|*(2−x^2)^{-1/2}] \] \[D[y]=D[|x|*(2−x^2)^{-1/2}] \] \[D[y]=D[|x|]*(2−x^2)^{-1/2}+|x|*D[(2−x^2)^{-1/2}] \] \[y'=\frac{x}{|x|}(2−x^2)^{-1/2}+\frac122x|x|(2−x^2)^{-3/2} \] \[y'=\frac{x}{|x|}(2−x^2)^{-1/2}+x|x|(2−x^2)^{-3/2} \] see if i worked that out correctly :)
amistre64
  • amistre64
quotient rule is often times a pain; it tends to be easier to turn up the denominator and just run thru the product rule
anonymous
  • anonymous
Thank you! that gives the slope I need to find my equation :)

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