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steel11
Group Title
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.
 one year ago
 one year ago
steel11 Group Title
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.
 one year ago
 one year ago

This Question is Closed

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

steel11 Group TitleBest ResponseYou've already chosen the best response.0
product rule? yes
 one year ago

uzumakhi Group TitleBest ResponseYou've already chosen the best response.0
try to use it
 one year ago

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

uzumakhi Group TitleBest ResponseYou've already chosen the best response.0
try to use my method and tell me the answer u got
 one year ago

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

steel11 Group TitleBest ResponseYou've already chosen the best response.0
thank you across. you used product rule and chain rule to solve right?
 one year ago

steel11 Group TitleBest ResponseYou've already chosen the best response.0
guess uzumaki was right on having to use product rule. ._.
 one year ago

uzumakhi Group TitleBest ResponseYou've already chosen the best response.0
that what i am talking about thanx @across
 one year ago

across Group TitleBest ResponseYou've already chosen the best response.1
Yes. 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\[f'(x)=2\cos\left(\frac1x\right)+2x\cos\left(\frac1x\right)\cdot\left(\frac1{x^2}\right).\]
 one year ago

across Group TitleBest ResponseYou've already chosen the best response.1
I forgot the negative sign in the last term, but you get the idea. I used the product and chain rules.
 one year ago
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