if y=cosx.sinx find dy/dx
just use the product rule
If y = fg, then
y' = f'g + fg'
So,
dy.dx = -sinx.sinx + cosx.cosx
= cos^2x - sin^2x
= cos2x
Or, you could have noticed that
y = 1/2 sin2x, so
y' = cos2x
To find the derivative of y with respect to x, dy/dx, we can use the product rule.
The product rule states that if we have two functions u(x) and v(x), then the derivative of their product with respect to x is given by:
(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
In this case, let's define u(x) = cos(x) and v(x) = sin(x).
Now, let's find the derivatives of u(x) and v(x):
u'(x) = -sin(x) (derivative of cos(x) is -sin(x))
v'(x) = cos(x) (derivative of sin(x) is cos(x))
Now, we can apply the product rule:
dy/dx = u'(x)v(x) + u(x)v'(x)
= (-sin(x))(sin(x)) + (cos(x))(cos(x))
= -sin^2(x) + cos^2(x)
Therefore, the derivative of y = cos(x)sin(x) with respect to x, dy/dx, is -sin^2(x) + cos^2(x).