How do you differentiate x*sin(2/x)?

1 Answer
Jun 20, 2017

(x*sin(2/x))' = sin(2/x)-(2cos(2/x))/x

Explanation:

We are going to have to use more than one rule here, first the product rule. If we were to look at just "x*sin(x)", you might know how to differentiate the trigonometric function sin(x) and the linear function x. The product rule states:

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

so (x*sin(x))' = 1*sin(x)+x*cos(x)

Similarly (x*sin(2/x))' = 1*sin(2/x)+x*(sin(2/x))'

Looking at (sin(2/x))', this require the use of the chain rule:
dy/dx = dy/(du)*(du)/dx

Let's say u = 2/x , so sin(2/x) = sin(u)

This means: d/(du)sin(u) = cos(u) and d/dx(2*x^-1)=-2x^-2

then we would get:
(sin(2/x))' = (sin(u))' * (2/x)' = cos(u) * (-2x^-2) = -(2cos(2/x))/x^2

Putting all this together, we get:

(x*sin(2/x))' = sin(2/x)+x*(-(2cos(2/x))/x^2) = sin(2/x)-(2cos(2/x))/x

PS. if you know how to differentiate and integrate sin(x) and cos(x), then you know more than me, because I can't remember those to safe my life :)