What is the derivative of #y = sin(sin(sinx))#?

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Answer 1 · 2015-06-05T12:25:12

We have #(sin(u))' = u'cos(u)#

where #u = sin(sin(x))#

#u' = (sin(sin(x)))' = cos(x)*cos(sin(x))#

So #(sin(u))' = cos(x)cos(sin(x))cos(sin(sin(x))#

Sorry if in your countries you need to use the "chain rules" in France we don't use that, we just remember that :

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