Question

How do you use the chain rule to differentiate `y=sin(cosx)`?

Answer 1

`=>dy/dx=-sin(x)cos(cos(x))`

Explanation:

Let `u=cos(x)=> (du)/dx = -sin(x)`

Let `y=sin(u)=> dy/(du)=cos(u)`

But `color(blue)(dy/(du)xx(du)/dx)color(red)(" " =" " (du)/(du)xxdy/dx)color(green)(" "=" "1xxdy/dx)color(magenta)(" "=" "dy/dx)`

`=>dy/dx=-sin(x)cos(u)`

but `u=cos(x)` giving:

`=>dy/dx=-sin(x)cos(cos(x))`