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

1 Answer

Nov 1, 2016

# dy/dx=-sinxcos(cosx)#

Explanation:

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)#

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