Y=sin X is a function because for each value of X, there is exactly one Y value.
D(y)= sin 2x
y=3 cos(x) y' = -3 sin(x)
That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]
2
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
Y=sin X is a function because for each value of X, there is exactly one Y value.
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
The amplitude of the wave [ y = -2 sin(x) ] is 2.
D(y)= sin 2x
The range of -sin x depends on the domain of x. If the domain of x is unrestricted then the range of y is [-1, 1].
It is zero.
y=3 cos(x) y' = -3 sin(x)
y=-10 sin 5x sin 5x=y/-10 x=asin(y/-10)/5
y = x sin(x) + cos(x)Derivative of the first term = x cos(x) + sin(x)Derivative of the second term = -sin(x)y' = Sum of the derivatives = x cos(x) + sin(x) - sin(x)= [ x cos(x) ]
5
The amplitude is 1.