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Find the differential of the function.

z = e⁻⁴ˣ cos 6πt

dz = (?)dx+(?)dt

User Jaquarh
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1 Answer

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Final answer:

To find the differential of the function z = e^{-4x} cos(6\pi t), differentiate partially with respect to both x and t using the product rule and chain rule, resulting in dz = -4e^{-4x} cos(6\pi t) dx - 6\pi e^{-4x} sin(6\pi t) dt.

Step-by-step explanation:

To find the differential of the function z = e^{-4x} \cos(6\pi t), we need to use the product rule and chain rule to differentiate with respect to both x and t.

The differential dz is composed of two parts: one for x and one for t. Calculating the partial derivative with respect to x, we get:

d( e^{-4x} \cos(6\pi t) )/dx = \cos(6\pi t) \cdot d( e^{-4x})/dx + e^{-4x} \cdot d(\cos(6\pi t))/dx

Since \cos(6\pi t) is not a function of x, its derivative with respect to x is 0, and we only need to differentiate e^{-4x}, which gives us -4e^{-4x} \cos(6\pi t). Then, we have:

\partial z/\partial x = -4e^{-4x} \cos(6\pi t)

Similarly, we find the partial derivative with respect to t:

d( e^{-4x} \cos(6\pi t) )/dt = e^{-4x} \cdot d(\cos(6\pi t))/dt + \cos(6\pi t) \cdot d( e^{-4x})/dt

Here, e^{-4x} is a constant with respect to t, so its derivative is 0, and we need to differentiate \cos(6\pi t) with respect to t, obtaining -6\pi e^{-4x} \sin(6\pi t). Thus:

\partial z/\partial t = -6\pi e^{-4x} \sin(6\pi t)

In conclusion, the differential dz for the function z is:

dz = -4e^{-4x} \cos(6\pi t) dx - 6\pi e^{-4x} \sin(6\pi t) dt

User Jeca
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