Question

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

Answer 1

Since x>0, then |x|=x

Answer 2

You need to use quotient rule + chain rule + product rule + constant rule

Answer 3

\[\sqrt{2-x^2}=(2-x^2)^\frac{1}{2}\]
\[[[(2-x^2)]^\frac{1}{2}]'=\frac{1}{2}(2-x^2)^{\frac{1}{2}-1}(2-x^2)' \text{ note: I applied chain rule here}\]

Answer 4

do i then need to apply the quotient rule using the absolute value of x?

Answer 5

\[y=|x|*(2−x^2)^{-1/2} \]
\[D[y=|x|*(2−x^2)^{-1/2}] \]
\[D[y]=D[|x|*(2−x^2)^{-1/2}] \]
\[D[y]=D[|x|]*(2−x^2)^{-1/2}+|x|*D[(2−x^2)^{-1/2}] \]
\[y'=\frac{x}{|x|}(2−x^2)^{-1/2}+\frac122x|x|(2−x^2)^{-3/2} \]
\[y'=\frac{x}{|x|}(2−x^2)^{-1/2}+x|x|(2−x^2)^{-3/2} \]
see if i worked that out correctly :)

Answer 6

quotient rule is often times a pain; it tends to be easier to turn up the denominator and just run thru the product rule

Answer 7

Thank you! that gives the slope I need to find my equation :)