How do you find the derivative of $y= x*sin(1/x)$ ?

Question

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Answer 1 · 2014-07-26T04:21:11

Answer is $y'=-1/x*cos(1/x)+sin(1/x)$

Solution

$y=f(x)*g(x)$

$y'=f(x)*g'(x)+f'(x)*g(x)$

Similarly, for the function mentioned in question,

$y'=x*(sin(1/x))'+sin(1/x)$

Now considering $v=sin(f(x))$

then,

$v'=(sinf(x))'=cos(f(x))*f'(x)$ ,

which implies,

$(sin(1/x))'=cos(1/x)(-1/x^2)$

Hence,

$y'=-1/x*cos(1/x)+sin(1/x)$

How do you find the derivative of y= x*sin(1/x)y= x*sin(1/x) ?