A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
derivative of 2xsin(1/x)
my answer i put, which is wrong
2sinx^1*cosx^1*1x^2
can anybody give me correct answer or find where i went wrong? im doing the chain rule, and i think you have to do twice, which i did.
btw, i brought the x to the numerator with 1 as exponent.
 2 years ago
derivative of 2xsin(1/x) my answer i put, which is wrong 2sinx^1*cosx^1*1x^2 can anybody give me correct answer or find where i went wrong? im doing the chain rule, and i think you have to do twice, which i did. btw, i brought the x to the numerator with 1 as exponent.

This Question is Closed

uzumakhi
 2 years ago
Best ResponseYou've already chosen the best response.0do u know abut product derivation d\dx(a*b) = a d\dx(b) + b d\dx(a)

steel11
 2 years ago
Best ResponseYou've already chosen the best response.0I would like to do this with chain rule. If thats possible, which I think it is.

uzumakhi
 2 years ago
Best ResponseYou've already chosen the best response.0try to use my method and tell me the answer u got

across
 2 years ago
Best ResponseYou've already chosen the best response.1Let\[f(x)=2x\sin\left(\frac1x\right).\]Then\[f'(x)=2\sin\left(\frac1x\right)+2x\cos\left(\frac1x\right)\cdot\left(\frac1{x^2}\right)=2\left[\sin\left(\frac1x\right)\frac1x\cos\left(\frac1x\right)\right].\]

steel11
 2 years ago
Best ResponseYou've already chosen the best response.0thank you across. you used product rule and chain rule to solve right?

steel11
 2 years ago
Best ResponseYou've already chosen the best response.0guess uzumaki was right on having to use product rule. ._.

uzumakhi
 2 years ago
Best ResponseYou've already chosen the best response.0that what i am talking about thanx @across

across
 2 years ago
Best ResponseYou've already chosen the best response.1Yes. To be more specific, let me break it down for you: Essentially, you have this:\[f(x)=g(x)h(j(x)),\]where\[g(x)=2x\\h(x)=\sin x\\j(x)=\frac1x.\]Then\[f'(x)=g'(x)h(j(x))+g(x)h'(j(x))j'(x),\]and since\[g'(x)=2\\h'(x)=\cos x\\j'(x)=\frac1{x^2},\]we have that\[f'(x)=2\cos\left(\frac1x\right)+2x\cos\left(\frac1x\right)\cdot\left(\frac1{x^2}\right).\]

across
 2 years ago
Best ResponseYou've already chosen the best response.1I forgot the negative sign in the last term, but you get the idea. I used the product and chain rules.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
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.