How do you find the equation of the tangent line to the curve #y=absx/sqrt(2-x^2)# at (1,1)?

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Answer 1 · 2016-11-29T01:58:37

Please see below.

For #x# near #1#, #x . 0#, so #absx = x# and #y = x/sqrt(2-x^2)#.

Using the quotient rule for derivatives and the chain rule to differentiate #sqrt(2-x^2)#, we get

#y' = ((1)(sqrt(2-x^2))-(x)(1/(2sqrt(2-x^2))*-2x))/(sqrt(2-x^2))^2#

# = (sqrt(2-x^2) +x^2/sqrt(2-x^2))/(2-x^2)#

We could rewrite some more, but we're really only interested in #y'# at #x=1#, so let's just do the arithmetic.

At #x=1#, the slope of the tangent line is

#m = (sqrt(2-(1)^2) +(1)^2/sqrt(2-(1)^2))/(2-(1)^2) = (1+1)/1 = 2#.

The equation of the line through #(1,1)# with slope #2# is

#y-1=2(x-1)# or

#y = 2x-1#.