Chapter 9: Problem 12
Find the derivative of each function. \(f(t)=\sqrt{3 t^{2}-t}\)
Short Answer
Expert verified
The derivative of the function \(f(t) = \sqrt{3t^2 - t}\) is:
\[f'(t) = \frac{6t - 1}{2\sqrt{3t^2 - t}}\]
Step by step solution
01
Identify the outer and inner functions
In this case, the outer function is the square root, which we can rewrite as a power of 1/2: \[f(t) = (3t^2 - t)^{1/2}\]
The inner function is the polynomial inside the square root: \(g(t) = 3t^2 - t\)
02
Find the derivatives of the outer and inner functions
First, find the derivative of the outer function with respect to its argument, using the power rule:
\[\frac{df(u)}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2}u^{-1/2}\]
where u is a placeholder for the inner function.
Next, find the derivative of the inner function with respect to t:
\[\frac{dg(t)}{dt} = \frac{d(3t^2 - t)}{dt} = 6t - 1\]
03
Use the chain rule to find the derivative of the original function
The chain rule states that the derivative of a composite function is the product of the derivative of the outer function and the derivative of the inner function, or in our case:
\[\frac{df(t)}{dt} = \frac{df(u)}{du} \times \frac{dg(t)}{dt}\]
04
Substitute the expressions found in Step 2 into the chain rule formula
Replace \(u\) with the inner function \(g(t)\), and substitute the derivatives found in Step 2:
\[\frac{df(t)}{dt} = \left(\frac{1}{2}(3t^2 - t)^{-1/2}\right) \times (6t - 1)\]
05
Simplify the final expression
Now, we need to simplify the expression to find the final derivative of the original function:
\[\frac{df(t)}{dt} = \frac{1}{2}(3t^2 - t)^{-1/2} \times (6t - 1) = \frac{6t - 1}{2\sqrt{3t^2 - t}}\]
The derivative of the function \(f(t) = \sqrt{3t^2 - t}\) is:
\[f'(t) = \frac{6t - 1}{2\sqrt{3t^2 - t}}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental concept in calculus, especially when you deal with composite functions. A composite function is essentially a function within another function. For instance, in our exercise, the function \(f(t)=\sqrt{3t^2-t}\) is composed of two functions: the outer square root function and the inner polynomial function \(3t^2-t\).
The chain rule helps us find the derivative of such composite functions efficiently. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function itself. In simple terms:
The chain rule helps us find the derivative of such composite functions efficiently. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function itself. In simple terms:
- Differentiate the outer function.
- Evaluate it with the inner function unchanged.
- Multiply this result by the derivative of the inner function.
Power Rule
The power rule is a straightforward and commonly used rule in calculus to find the derivative of functions expressed as powers of \(x\). It states that if you have a function \(f(x) = x^n\), the derivative will be \(f'(x) = nx^{n-1}\). In the exercise, we initially rewrote the square root function as a power: \(f(t) = (3t^2 - t)^{1/2}\). By doing so, applying the power rule becomes more apparent, and it aids in the stepwise differentiation process.
- Determine the power of the expression.
- Multiply by the power.
- Decrease the power by one.
Outer and Inner Functions
Identifying outer and inner functions is a critical step when using the chain rule. It involves a keen look at the structure of your function to separate it into layers.
In the provided exercise, we define:
In the provided exercise, we define:
- The outer function as the square root, \(f(u) = u^{1/2}\), after rewriting the original square root as a power for easier differentiation.
- The inner function is the polynomial \(g(t) = 3t^2 - t\) that resides within the outer square root function.