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

Jul 26, 2014

Answer is $y ' = - \frac{1}{x} \cdot \cos \left(\frac{1}{x}\right) + \sin \left(\frac{1}{x}\right)$

Solution

Using the product rule,

$y = f \left(x\right) \cdot g \left(x\right)$

$y ' = f \left(x\right) \cdot g ' \left(x\right) + f ' \left(x\right) \cdot g \left(x\right)$

Similarly, for the function mentioned in question,

$y ' = x \cdot \left(\sin \left(\frac{1}{x}\right)\right) ' + \sin \left(\frac{1}{x}\right)$

Now considering $v = \sin \left(f \left(x\right)\right)$

then,

$v ' = \left(\sin f \left(x\right)\right) ' = \cos \left(f \left(x\right)\right) \cdot f ' \left(x\right)$,

which implies,

$\left(\sin \left(\frac{1}{x}\right)\right) ' = \cos \left(\frac{1}{x}\right) \left(- \frac{1}{x} ^ 2\right)$

Hence,

$y ' = - \frac{1}{x} \cdot \cos \left(\frac{1}{x}\right) + \sin \left(\frac{1}{x}\right)$