What is the derivative of $sin(cosx)$ ?

Question

7747 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-01T21:08:37

$dy/dx=-sinxcos(cosx)$

You need to use the chain rule,
$d/dxf(g(x)) =f'(g(x))g'(x)$ or, $dy/dx=dy/(du)(du)/dx$

So, if $y=sin(cosx) => dy/dx=cos(cosx)d/dxcosx$
$:, dy/dx=cos(cosx)(-sinx)$
$:. dy/dx=-sinxcos(cosx)$

What is the derivative of sin(cosx)sin(cosx)?