solve calculus math problem?

y ' = (x) ' sin(1/x) + x [sin(1/x)] '

= (1) sin(1/x) + x [cos(1/x)]

= sin (1/x) + xcos(1/x)

P.S. I aint' sure if ya have to differentiate the (cos (1/x)) but that's about as far as I can get.

Good Luck!

You must take into account the fact that sin(1/x) and cos(1/x) are both functions of a function and therefore aren't just simple differentiations.

sin(1/x)-x*cos(1/x)(1/x^2)

I think that is correct using the product rule then the chain rule on the last part. "Derivative of x times sin(1/x) plus x times the derivative of sin(1/x)"

First rewrite the problem without a fraction.

y= x∙sin(x^-1)

You need to use the product rule.

Hi d Ho plus Ho d Hi

Hi = x

Ho = sin(1/x)

y'= x∙cos(x^-1)(x^-2) + sin(x^-1)(1)

y'=(x^-1)∙cos(x^-1) + sin(x^-1)

y'=[cos(1/x)]/x + sin(1/x)

sin(1/x) -1/(x^2) * cos(1/x) * x

not differancialble at thepoint x=0

f (x) = x . sin (1/x)

f ` (x) = 1. sin (1/x) + cos (1/x) . (-1/x²)

f ` (x) = sin (1/x) - (1/x²).cos(1/x)

dy/dx=-xcos(1/x)(-1/x^2)+1 sin (1/x)

dy/dx=(cos (1/x))/x+ sin (1/x)

d/dx(x sin (1/x))

Use the product rule

d/dx(uv)= du/dx (v)+(u)dv/dx

where u=x and v= sin(1/x)

= x. d/dx(sin(1/x)) +sin(1/x).d/dx (x)

= x. d/dx(sin(1/x)) +sin(1/x).

=x. cos(1/x) d/dx(1/x)+sin(1/x)

= sin(1/x) - (cos(1/x) / x)

dy/dx=sin(1/x)+x[cos(1/x)](-1/x^2)

Can you solve the problem?