By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
Differentials represent the instantaneous change of a variable. As the given function has only one variable, the differential can be found by using ordinary derivatives. It follows:
dy = y'(x) · dx (1)
If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:
[tex]y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x[/tex]
[tex]y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}[/tex]
[tex]dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx[/tex]
[tex]dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)[/tex]
[tex]dy = \frac{1}{2\pi}[/tex]
By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
To learn more on differentials: https://brainly.com/question/24062595
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