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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

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Since x>0, then |x|=x
You need to use quotient rule + chain rule + product rule + constant rule
\[\sqrt{2-x^2}=(2-x^2)^\frac{1}{2}\] \[[[(2-x^2)]^\frac{1}{2}]'=\frac{1}{2}(2-x^2)^{\frac{1}{2}-1}(2-x^2)' \text{ note: I applied chain rule here}\]

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

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