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anonymous
 3 years 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
anonymous
 3 years 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

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myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Since x>0, then x=x

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1You need to use quotient rule + chain rule + product rule + constant rule

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do i then need to apply the quotient rule using the absolute value of x?

amistre64
 3 years ago
Best 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 :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2quotient rule is often times a pain; it tends to be easier to turn up the denominator and just run thru the product rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you! that gives the slope I need to find my equation :)
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