Question

The Derivative of sin 2\(x\) Graph the function \(y=2 \cos 2 x\) for \(-2 \leq x \leq 3.5 .\) Then, on the same screen, graph $$\quad y=\frac{\sin 2(x+h)-\sin 2 x}{h}$$ for \(h=1.0,0.5,\) and \(0.2 .\) Experiment with other values of \(h,\) including negative values. What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.

Short answer

As \(h \rightarrow 0\), the function \(y=\frac{\sin 2(x+h)-\sin 2 x}{h}\) starts to resemble the function \(y = 2 \cos 2x\) plotted initially. This is because as the value of \(h\) decreases, it's getting replaced with \(x\), so our function is getting closer to the derivative of the original function \(y=2 \cos 2 x\).

Step-by-step solution

Graphing the Function \(y=2 \cos 2 x\)

Begin by plotting \(y=2 \cos 2 x\) in the given range \(x \in [-2, 3.5]\). Use any graphing tool to draw the curve. This curve of cosine function will be the reference for comparing with other curves.

Plotting for various values of h

Then plot the function \(y=\frac{\sin 2(x+h)-\sin 2 x}{h}\) for different values of \(h (1.0, 0.5, 0.2)\). Each of these functions will have a different curve on the graph. Compare these plotted curves with the curve from Step 1 to understand the changes when using different values of h.

Experiment with other values of h

Now, try other values of \(h\) including negative ones and observe the changes to the graph. Make a note of how the curve changes with each variation of \(h\).

Analyze as \(h \rightarrow 0\)

As \(h \rightarrow 0\), the function \(y=\frac{\sin 2(x+h)-\sin 2 x}{h}\) starts to resemble the function \(y = 2 \cos 2x\) plotted initially. This is because as the value of \(h\) decreases, it is getting replaced with \(x\), so our function is getting closer to the derivative of the original function \(y=2 \cos 2 x\).