## 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

hi uzu :)

2. uzumakhi Group Title

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

3. steel11 Group Title

product rule? yes

4. uzumakhi Group Title

try to use it

5. steel11 Group Title

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

6. uzumakhi Group Title

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

7. across Group Title

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

8. steel11 Group Title

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

9. steel11 Group Title

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

10. uzumakhi Group Title

that what i am talking about thanx @across

11. steel11 Group Title

=]

12. across Group Title

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

13. across Group Title

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