What is the derivative of $cos(pi/x)/x^2$ ?

Question

1605 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-12T16:29:32

Quotient Rule
$d/(dx)[f(x)/(g(x))] = (g(x)f'(x) - f(x)g'(x))/(g(x))^2$

$f(x) = cos(pi/x)$
$g(x) = x^2$

The derivatives in general are:
$d/(dx)[cosu] = -sinu((du)/(dx))$
$d/(dx)[u^2] = 2u((du)/(dx))$

$=> [cancel(x^2)[-sin(pi/x)overbrace((pi)(-1/cancel(x^2)))^("Chain Rule")] - cos(pi/x)*2x]/(x^2)^2$

Simplify:
$= [pisin(pi/x) - 2xcos(pi/x)]/x^4$

What is the derivative of cos(pi/x)/x^2 cos(pi/x)/x^2 ?