What is the derivative of y = x sin (5/x)?

Sep 25, 2016

Let $g \left(x\right) = x \mathmr{and} h \left(x\right) = \sin \left(\frac{5}{x}\right)$ and then use the product rule but we will need the chain rule to obtain h'(x)

Explanation:

The product rule is:

$\left\{g \left(x\right) h \left(x\right)\right\} ' = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

$L e t g \left(x\right) = x \mathmr{and} h \left(x\right) = \sin \left(\frac{5}{x}\right)$

$g ' \left(x\right) = 1$

To apply the chain rule on h'(x), we let $h \left(x\right) = h \left(u \left(x\right)\right)$
where $h \left(u\right) = \sin \left(u\right)$ and u(x) = 5/x

$h ' \left(u\right) = \cos \left(u\right)$
u'(x) = -5/(x²)

The chain rule says to multiply u'(x) and h'(u):

(-5/(x²))cos(u)

Reversing the u substitution:

h'(x) = (-5/(x²))cos(5/x)

Substituting the above into the product rule:

$d \frac{x \sin \left(\frac{5}{x}\right)}{\mathrm{dx}} = \sin \left(\frac{5}{x}\right) - 5 \cos \frac{\frac{5}{x}}{x}$