How do you find the derivative of #y= sin(xcosx)#?

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Answer 1 · 2015-05-24T03:24:01

There are two ways I can think of to do this. Substitution, or just thinking as you go.

1)
If you want to make this look nicer, you can write this as:

#y = sin(xcosx) = sin(u)#
where #u = u(x) = xcosx#.

#d/(du)[sinu(x)] = cosu*((du)/(dx))#

Then,
#(du)/(dx) = d/(dx)[xcosx] = x*d/(dx)[cosx] + cosx*d/(dx)[x]#

# = -xsinx + cosx#

So, substituting that back in, we get:

#(dy)/(dx) = d/(dx)[sin(xcosx)] = cos(xcosx)*(-xsinx + cosx)#

#= -cos(xcosx)*(xsinx - cosx)#

2)

Or, you can just take the derivative and do the chain rule as you go.

#d/(dx)[sin(xcosx)] = cos(xcosx)*[x*-sinx + cosx*1]#

#= -cos(xcosx)*(xsinx - cosx)#