Find the differential of the function.
$$
z=e^{-2 x} \cos 2 \pi t
$$

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We need to find the differential of the function $z = e^{-2x} \cos(2\pi t)$. ‖ ‖ step 2 ⋮ To find the differential, we use the product rule for differentiation. The product rule states that if $u$ and $v$ are functions of $x$ and $t$, then $d(uv) = u \, dv + v \, du$. Here, $u = e^{-2x}$ and $v = \cos(2\pi t)$. ‖ ‖ step 3 ⋮ First, we differentiate $u = e^{-2x}$ with respect to $x$: $$ \frac{du}{dx} = \frac{d}{dx}(e^{-2x}) = -2e^{-2x} $$ ‖ ‖ step 4 ⋮ Next, we differentiate $v = \cos(2\pi t)$ with respect to $t$: $$ \frac{dv}{dt} = \frac{d}{dt}(\cos(2\pi t)) = -2\pi \sin(2\pi t) $$ ‖ ‖ step 5 ⋮ Now, applying the product rule: $$ dz = e^{-2x} \, d(\cos(2\pi t)) + \cos(2\pi t) \, d(e^{-2x}) $$ ‖ ‖ step 6 ⋮ Substituting the derivatives we found: $$ dz = e^{-2x} (-2\pi \sin(2\pi t) \, dt) + \cos(2\pi t) (-2e^{-2x} \, dx) $$ ‖ ‖ step 7 ⋮ Simplifying the expression: $$ dz = -2\pi e^{-2x} \sin(2\pi t) \, dt - 2e^{-2x} \cos(2\pi t) \, dx $$ ‖ ∻Answer∻ ⚹ The differential of the function $z = e^{-2x} \cos(2\pi t)$ is $dz = -2\pi e^{-2x} \sin(2\pi t) \, dt - 2e^{-2x} \cos(2\pi t) \, dx$. ⚹ ∻Key Concept∻ ⚹Differentiation using the product rule⚹ ∻Explanation∻ ⚹The product rule is used to differentiate a product of two functions. In this case, we differentiated $e^{-2x}$ and $\cos(2\pi t)$ separately and then applied the product rule to find the differential of the given function.⚹◊What is the differential of the function $z = e^{-2x} \cos (2\pi t)$⍭ Generate me a similar question◊

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