Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
can someone help me simplify the derivative of x/sqrt(1x^2) ? This is for problem 5A3G of the second problem sets, and I think my proficiency with radicals may not be up to snuff to be able to properly solve this.
 one year ago
 one year ago
can someone help me simplify the derivative of x/sqrt(1x^2) ? This is for problem 5A3G of the second problem sets, and I think my proficiency with radicals may not be up to snuff to be able to properly solve this.
 one year ago
 one year ago

This Question is Closed

TopiBest ResponseYou've already chosen the best response.1
What do you mean by simplifing? Let's calculate the derivative first:\[\frac{ d }{ dx }\frac{ x }{ \sqrt{1x ^{2}} }=\frac{ d }{ dx }x(1x ^{2})^{\frac{ 1 }{ 2 }}=x \frac{ d }{ dx }(1x ^{2})^{\frac{ 1 }{ 2 }}+(1x ^{2})^{\frac{ 1 }{ 2 }}\]\[=x(\frac{ 1 }{ 2 }(1x ^{2})^{\frac{ 3 }{ 2 }})\frac{ d }{dx }(1x ^{2})+(1x^2)^{\frac{ 1 }{ 2 }}\]\[=\frac{ 1 }{ 2 } x(1x^2)^{\frac{ 3 }{ 2 }}(2x)+(1x^2)^{\frac{ 1 }{ 2 }}=\frac{ x^2 }{ \sqrt{(1x^2)^3} }+\frac{ 1 }{ \sqrt{1x^2} }\]If you want to take that over common denominator, you have to multiply the last term with\[\frac{ 1x^2 }{ 1x^2}\]so you get\[\frac{ x^2+1x^2 }{ \sqrt{(1x^2)^3} } = \frac{ 1 }{ \sqrt{(1x^2)^3} }\]That might be the simplest form for that derivative.
 one year ago

lau22Best ResponseYou've already chosen the best response.0
thank you very much! that's exactly what I needed!
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.