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Find the derivative of each function. \(f(t)=\left(3 t^{2}-2 t+1\right)^{3 / 2}\)

Short Answer

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The derivative of the given function is: \[\frac{d}{dt}[(3t^2 - 2t + 1)^{\frac{3}{2}}] = \left(\frac{3}{2}(3t^2 - 2t + 1)^{\frac{1}{2}}\right)(6t - 2).\]
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Step 1: Find the derivative of f(u)

Since \(f(u) = u^{\frac{3}{2}}\), we can find the derivative of this function with respect to \(u\) using the power rule: \(\frac{d}{du}[u^n] = nu^{(n-1)}\). In this case, \(n = \frac{3}{2}\). Applying the power rule, we find the derivative of \(f(u)\) to be: \[\frac{df}{du} = \frac{3}{2}u^{\frac{1}{2}}\].

Step 2: Find the derivative of g(t)

The function \(g(t) = 3t^2 - 2t + 1\) is a quadratic function. We can find the derivative with respect to \(t\) using the power rule for each term: \[\frac{dg}{dt} = \frac{d}{dt}[3t^2 - 2t + 1] = \frac{d}{dt}(3t^2) - \frac{d}{dt}(2t) + \frac{d}{dt}(1).\] Now, apply the power rule to each term: \[\frac{dg}{dt} = 6t - 2.\]

Step 3: Apply the chain rule

Now, we'll use the chain rule to find the overall derivative. We have: \[\frac{d}{dt}[(3t^2 - 2t + 1)^{\frac{3}{2}}] = \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = \frac{df}{du} \cdot \frac{dg}{dt}.\] Substitute the derivatives we found in Steps 1 and 2: \[\frac{dy}{dt} = \left(\frac{3}{2}u^{\frac{1}{2}}\right) \cdot (6t - 2).\] Finally, replace \(u\) with \(g(t) = 3t^2 - 2t + 1\): \[\frac{dy}{dt} = \left(\frac{3}{2}(3t^2 - 2t + 1)^{\frac{1}{2}}\right)(6t - 2).\] This is the derivative of the given function: \[\frac{d}{dt}[(3t^2 - 2t + 1)^{\frac{3}{2}}] = \left(\frac{3}{2}(3t^2 - 2t + 1)^{\frac{1}{2}}\right)(6t - 2).\]

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