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

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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)#