# What is derivative of function sin(1+x)*cos(1-x)?

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We'll use the product rule to determine the derivative of this function:

(u*v)' = u'*v + u*v'

We'll use the chain rule to differentiate the terms of the product.

Let u = sin(1+x) = > u' = cos(1+x)

Let v = cos (1-x) => v' = -(-1)*sin(1 - x)

The derivative of the function is:

y' = cos(1+x)*cos(1-x) + sin(1+x)*sin(1-x)

We'll transform the product of trigonometric functions into products:

cos(1+x)*cos(1-x) = (1/2)*[cos(1+x+1-x) + cos(1+x-1+x)]

cos(1+x)*cos(1-x) = (cos 2 + cos (2x))/2

sin(1+x)*sin(1-x) = (1/2)*[cos(1+x-1+x) - cos(1+x+1-x)]

sin(1+x)*sin(1-x) = (cos (2x) - cos 2)/2

y' = [cos 2 + cos (2x) + cos (2x) - cos 2]/2

y' = cos 2x

**The derivative of the given function is: y' = cos 2x**