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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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 :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question