deriv.
y=x(sinx)(cosx)
y=uvw
y'=vwu' + uvw'+uwv'
It is just algebra, and probably some trig identities.
use the product rule.
hard way:
(1)[(sinx)(cosx)] + {x(cosx)](cosx) + [x(sinx)](-sinx)
= sinx cosx + xcos^2 x - xsin^2 x
= 1/2 sin 2x + xcos 2x
easier way:
y = 1/2 x sin 2x
y' = 1/2 (sin 2x) + 1/2 x (2cos 2x)
= 1/2 sin 2x + x cos 2x
To find the derivative of the given function y = x*sin(x)*cos(x), you can use the product rule and the chain rule.
The product rule states that if you have two functions u(x) and v(x), the derivative of their product f(x) = u(x)*v(x) is given by:
f'(x) = u'(x)*v(x) + u(x)*v'(x)
Let's apply the product rule to the given function.
1. Identify the two functions:
u(x) = x*sin(x)
v(x) = cos(x)
2. Calculate the derivatives of the two functions:
u'(x) = (1*cos(x)) + (x*cos(x)) [Applying the product rule and the chain rule]
= cos(x) + x*cos(x)
v'(x) = -sin(x)
3. Apply the product rule formula:
y' = u'(x)*v(x) + u(x)*v'(x)
= (cos(x) + x*cos(x)) * cos(x) + x*sin(x) * (-sin(x)).
4. Simplify the expression:
y' = cos^2(x) + x*cos(x)*cos(x) - x*sin^2(x)
So, the derivative of y = x*sin(x)*cos(x) is y' = cos^2(x) + x*cos(x)*cos(x) - x*sin^2(x).