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Find an equation of the tangent line to the bulletnose 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
 one year ago
 one year ago
Find an equation of the tangent line to the bulletnose 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
 one year ago
 one year ago

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myininayaBest ResponseYou've already chosen the best response.1
Since x>0, then x=x
 one year ago

myininayaBest ResponseYou've already chosen the best response.1
You need to use quotient rule + chain rule + product rule + constant rule
 one year ago

myininayaBest ResponseYou've already chosen the best response.1
\[\sqrt{2x^2}=(2x^2)^\frac{1}{2}\] \[[[(2x^2)]^\frac{1}{2}]'=\frac{1}{2}(2x^2)^{\frac{1}{2}1}(2x^2)' \text{ note: I applied chain rule here}\]
 one year ago

darkmareBest ResponseYou've already chosen the best response.0
do i then need to apply the quotient rule using the absolute value of x?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
\[y=x*(2−x^2)^{1/2} \] \[D[y=x*(2−x^2)^{1/2}] \] \[D[y]=D[x*(2−x^2)^{1/2}] \] \[D[y]=D[x]*(2−x^2)^{1/2}+x*D[(2−x^2)^{1/2}] \] \[y'=\frac{x}{x}(2−x^2)^{1/2}+\frac122xx(2−x^2)^{3/2} \] \[y'=\frac{x}{x}(2−x^2)^{1/2}+xx(2−x^2)^{3/2} \] see if i worked that out correctly :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
quotient rule is often times a pain; it tends to be easier to turn up the denominator and just run thru the product rule
 one year ago

darkmareBest ResponseYou've already chosen the best response.0
Thank you! that gives the slope I need to find my equation :)
 one year ago
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