Glissando said:
micromass said:Hi Glissando!
Your derivative is wrong. If you want to differentiate a product, then you can't just differentiate both factors and be done with it. You need to apply the product rule:
[tex](fg)^\prime=f^\prime g+fg^\prime[/tex]
The graph of y = sinx(1-cosx) is a sinusoidal curve that oscillates between positive and negative values. It has a period of 2π and an amplitude of 1.
The x-intercepts of the graph occur when sinx(1-cosx) = 0. This happens when either sinx = 0 or 1-cosx = 0. Solving for x in these equations, we get x = 0, π, 2π, and so on.
The maximum value occurs when sinx = 1 and 1-cosx = 1, which happens at x = π/2 and all multiples of 2π. The minimum value occurs when sinx = -1 and 1-cosx = 1, which happens at x = 3π/2 and all odd multiples of π.
Yes, you can sketch the graph using the properties of the sine and cosine functions. First, plot the x-intercepts at x = 0, π, 2π, etc. Then, plot the maximum and minimum points at x = π/2, 3π/2, 5π/2, etc. Finally, connect these points with a smooth curve to complete the graph.
The graph of y = sinx(1-cosx) is similar to the graph of y = sinx, but it is stretched vertically by a factor of 1-cosx. This means that the amplitude of y = sinx(1-cosx) is greater than the amplitude of y = sinx. Additionally, the x-intercepts and maximum and minimum points of y = sinx are also present in the graph of y = sinx(1-cosx).