Question

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.

Answer 1

hi uzu :)

Answer 2

do u know abut product derivation
d\dx(a*b) = a d\dx(b) + b d\dx(a)

Answer 3

product rule? yes

Answer 4

try to use it

Answer 5

I would like to do this with chain rule. If thats possible, which I think it is.

Answer 6

try to use my method
and tell me the answer u got

Answer 7

Let\[f(x)=2x\sin\left(\frac1x\right).\]Then\[f'(x)=2\sin\left(\frac1x\right)+2x\cos\left(\frac1x\right)\cdot\left(-\frac1{x^2}\right)=2\left[\sin\left(\frac1x\right)-\frac1x\cos\left(\frac1x\right)\right].\]

Answer 8

thank you across. you used product rule and chain rule to solve right?

Answer 9

guess uzumaki was right on having to use product rule. ._.

Answer 10

that what i am talking about thanx @across

Answer 11

=]

Answer 12

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\[f'(x)=2\cos\left(\frac1x\right)+2x\cos\left(\frac1x\right)\cdot\left(\frac1{x^2}\right).\]

Answer 13

I forgot the negative sign in the last term, but you get the idea. I used the product and chain rules.