How do you find the derivative of # y = sin(x cos x)# using the chain rule?

You will need to use the product rule to find #d/dx(xcosx)#, and then the chain rule to find #d/dxsin(xcos)#, so I will explain both;

Use of the Product Rule
If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

# d/dx(uv)=u(dv)/dx+v(du)/dx #, or, # (uv)' = (du)v + u(dv) #

I was taught to remember the rule in words; "The first times the derivative of the second plus the second times the derivative of the first ".

So with # xcosx # we have;

# d/dx(xcosx) =(x)(d/dxcosx) + (cosx)(d/dxx) #
# :. d/dx(xcosx) = (x)(-sinx) + (cosx)(1) #
# :. d/dx(xcosx) = cosx-xsinx # .... [1]

Use of the Chain Rule
You should learn the Chain Rule for Differentiation, and practice how to use it:

If # y=f(x) # then # f'(x)=dy/dx=dy/(du)(du)/dx #

I was taught to remember that the differential can be treated like a fraction and that the "#dx#'s" of a common variable will "cancel" (It is important to realise that #dy/dx# isn't a fraction but an operator that acts on a function, there is no such thing as "#dx#" or "#dy#" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

# dy/dx = dy/(dv)(dv)/(du)(du)/dx # etc, or # (dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx) #

So, If #y=sin(xcosx)#, Then:

# { ("Let "u=xcosx, => , (du)/dx=cosx-xsinx " from [1]"), ("Then "y=sinu, =>, dy/(du)=cosu ) :}#

Using # dy/dx=(dy/(du))((du)/dx) # we get:

# dy/dx=(cosu)(cosx-xsinx) #
# :. dy/dx=(cosx-xsinx)*cos(xcosx) #