How do you find the derivative of:
y= 2sinx cosx
You probably did it correctly but did not match the answer in the book.
Can be done in two ways:
1.
dy/dx = 2sinx(-cosx) + 2cosxcox
= 2cos^2 x - 2 sin^2 x
2. recognize that 2sinxcosx = sin 2x
so y = 2sinxcox
= sin 2x
dy/dx = 2cosx
but remember cos 2x = cos^2 x - sin^2 x
so the two answers are the same
You need to use the multiplication rule:
y = 2 sinx cosx
y' = 2 d/dx (sinx * cosx)
y' = 2 [ (sinx * -sinx) + (cosx * cosx) ]
y' = 2 [ -sin^2 x + cos^2 x ]
y' = 2cos^2 x - 2sin^2 x
To find the derivative of y = 2sin(x)cos(x), we can use the product rule of differentiation.
The product rule states that for two functions u(x) and v(x), the derivative of their product is given by:
d/dx (u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
In this case, u(x) = 2sin(x) and v(x) = cos(x).
First, let's find u'(x) and v'(x):
u'(x) = d/dx (2sin(x))
To differentiate sin(x), we use the chain rule, which states that d/dx (sin(x)) = cos(x). Therefore, u'(x) = 2cos(x).
v'(x) = d/dx (cos(x))
The derivative of cos(x) is given by d/dx (cos(x)) = -sin(x). Therefore, v'(x) = -sin(x).
Now, we can substitute our results into the product rule equation:
d/dx (2sin(x)cos(x)) = (2cos(x))(cos(x)) + (2sin(x))(-sin(x))
Simplifying this expression further, we get:
d/dx (2sin(x)cos(x)) = 2cos^2(x) - 2sin^2(x)
So, the derivative of y with respect to x is given by:
dy/dx = 2cos^2(x) - 2sin^2(x)