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

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Answer 1 · 2016-09-25T02:17:40

Let #g(x) = x and h(x) = sin(5/x)# and then use the product rule but we will need the chain rule to obtain h'(x)

The product rule is:

#{g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)#

#Let g(x) = x and h(x) = sin(5/x)#

#g'(x) = 1#

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

#h'(u) = cos(u)#
#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(xsin(5/x))/dx = sin(5/x) - 5cos(5/x)/x #