y=sin(2x)/x, x=Pi, dx=.25
find the differentiate
dydx=2cos(2x)x−sin(2x)x2.\frac{dy}{dx}=\frac{2cos(2x)}{x}-\frac{sin(2x)}{x^2}.dxdy=x2cos(2x)−x2sin(2x).
dy=[2cos(2x)x−sin(2x)x2]dx.dy=[\frac{2cos(2x)}{x}-\frac{sin(2x)}{x^2}]dx.dy=[x2cos(2x)−x2sin(2x)]dx.
When x=π, dx=0.25:x=\pi, \;dx=0.25:x=π,dx=0.25:
dy=[2cos(2π)π−sin(2π)π2]∗0.25=0.5π≈0.1592.dy=[\frac{2cos(2\pi)}{\pi}-\frac{sin(2\pi)}{\pi^2}]*0.25=\frac{0.5}{\pi}\approx 0.1592.dy=[π2cos(2π)−π2sin(2π)]∗0.25=π0.5≈0.1592.
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