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Problem 10

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

Expert verified
The derivative of the given function is: $\frac{d}{dt}[(3t^2 - 2t + 1)^{\frac{3}{2}}] = \left(\frac{3}{2}(3t^2 - 2t + 1)^{\frac{1}{2}}\right)(6t - 2).$
See the step by step solution

## 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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