Find the rate of change of the following functions: (a) \(\frac{3 t^{3}-t^{2}}{2 t}\) (b) \(\ln \sqrt{x}\) (c) \((t+2)(2 t-1)\) (d) \(\mathrm{e}^{3 v}\left(1-\mathrm{e}^{v}\right)\) (e) \(\sqrt{x}(\sqrt{x}-1)\)

To find the derivative of this function, we will use the quotient rule. Let \(u = 3t^3 - t^2\) and \(v = 2t\). The quotient rule states that for any two functions \(u(t)\) and \(v(t)\), the derivative of their quotient is: \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\) Now, we need to find the derivatives of the functions \(u(t)\) and \(v(t)\): \(u'(t) = \frac{d}{dt}(3t^3 - t^2) = 9t^2 - 2t\) \(v'(t) = \frac{d}{dt}(2t) = 2\) Substituting the derivatives into the quotient rule formula: \((\frac{u}{v})' = \frac{(9t^2 - 2t)(2t) - (3t^3 - t^2)(2)}{(2t)^2} = \frac{18t^3 - 4t^3}{4t^2} = \frac{(14t^3 - 4t^3)}{4t^2} = \frac{10t^3}{4t^2} = \frac{5t}{2}\) So, the rate of change of the function \(\frac{3 t^{3}-t^{2}}{2 t}\) is \(\frac{5t}{2}\).

First, rewrite the function as \(\ln x^{\frac{1}{2}}\). To find the derivative of this function, we will use the chain rule. Let \(u = x^{\frac{1}{2}}\). The chain rule states that for any two functions \(u(x)\) and \(g(u)\), the derivative of the composite function \(g(u(x))\) with respect to x is: \(g'(u(x)) = g'(u)u'(x)\) Now, we need to find the derivatives of the functions \(u(x)\) and \(g(u) = \ln u\): \(u'(x) = \frac{d}{dx}(x^{\frac{1}{2}}) = \frac{1}{2}x^{-\frac{1}{2}}\) \(g'(u) = \frac{d}{du}(\ln u) = \frac{1}{u}\) Substituting the derivatives into the chain rule formula: \(g'(u(x)) = (\frac{1}{u})(\frac{1}{2}x^{-\frac{1}{2}}) = \frac{1}{2x^{\frac{1}{2}}}\) So, the rate of change of the function \(\ln \sqrt{x}\) is \(\frac{1}{2x^{\frac{1}{2}}}\).

To find the derivative of this function, we will use the product rule. Let \(u = t+2\) and \(v = 2t-1\). The product rule states that for any two functions \(u(t)\) and \(v(t)\), the derivative of their product is: \((uv)' = u'v + uv'\) Now, we need to find the derivatives of the functions \(u(t)\) and \(v(t)\): \(u'(t) = \frac{d}{dt}(t+2) = 1\) \(v'(t) = \frac{d}{dt}(2t-1) = 2\) Substituting the derivatives into the product rule formula: \((uv)' = (1)(2t-1) + (t+2)(2) = 2t - 1 + 2t + 4 = 4t + 3\) So, the rate of change of the function \((t+2)(2 t-1)\) is \(4t + 3\).

To find the derivative of this function, we will use the product rule. Let \(u = \mathrm{e}^{3v}\) and \(v = (1-\mathrm{e}^v)\). The product rule states that for any two functions \(u(v)\) and \(v(v)\), the derivative of their product is: \((uv)' = u'v + uv'\) Now, we need to find the derivatives of the functions \(u(v)\) and \(v(v)\): \(u'(v) = \frac{d}{dv}(\mathrm{e}^{3v}) = 3\mathrm{e}^{3v}\) \(v'(v) = \frac{d}{dv}(1-\mathrm{e}^{v}) = -\mathrm{e}^{v}\) Substituting the derivatives into the product rule formula: \((uv)' = (3\mathrm{e}^{3v})(1-\mathrm{e}^v) + (\mathrm{e}^{3v})(-\mathrm{e}^v) = 3\mathrm{e}^{3v} - 3\mathrm{e}^{4v} - \mathrm{e}^{4v} = 3\mathrm{e}^{3v} - 4\mathrm{e}^{4v}\) So, the rate of change of the function \(\mathrm{e}^{3v}(1-\mathrm{e}^{v})\) is \(3\mathrm{e}^{3v} - 4\mathrm{e}^{4v}\).

To find the derivative of this function, we will use the product rule. Let \(u = x^{\frac{1}{2}}\) and \(v = x^{\frac{1}{2}} - 1\). The product rule states that for any two functions \(u(x)\) and \(v(x)\), the derivative of their product is: \((uv)' = u'v + uv'\) Now, we need to find the derivatives of the functions \(u(x)\) and \(v(x)\): \(u'(x) = \frac{d}{dx}(x^{\frac{1}{2}}) = \frac{1}{2}x^{-\frac{1}{2}}\) \(v'(x) = \frac{d}{dx}(x^{\frac{1}{2}} - 1) = \frac{1}{2}x^{-\frac{1}{2}}\) Substituting the derivatives into the product rule formula: \((uv)' = (\frac{1}{2}x^{-\frac{1}{2}})(x^{\frac{1}{2}} - 1) + (x^{\frac{1}{2}})(\frac{1}{2}x^{-\frac{1}{2}}) = \frac{1}{2}(x^{\frac{1}{2}} - 1) + \frac{1}{2} = \frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2} + \frac{1}{2} = \frac{1}{2}x^{\frac{1}{2}}\) So, the rate of change of the function \(\sqrt{x}(\sqrt{x}-1)\) is \(\frac{1}{2}x^{\frac{1}{2}}\).