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Chapter 5: Problem 48

Find the derivative of the function. $$ f(t)=t^{3 / 2} \log _{2} \sqrt{t+1} $$

Short Answer

Expert verified
Therefore, the derivative of the given function \(f(t)=t^{3 / 2} \log _{2} \sqrt{t+1} \) is \(f'(t) = \frac{3}{2}t^{1 / 2} \log _{2} \sqrt{t+1} + \frac{t^{3 / 2}}{2\ln2(t+1)}\).

Step by step solution

01

Identify the Functions

We first identify the functions that composed the original function. Here we have \(f_1(t) = t^{3 / 2}\) and \(f_2(t) = \log _{2} \sqrt{t+1}\).
02

Apply the Product Rule

Apply the product rule which states that the derivative of two functions multiplied together is equal to the derivative of the first function times the second function plus the first function times the derivative of the second function. Mathematically, the product rule is expressed as: \((f_1 f_2)' = f_1' f_2 + f_1 f_2'\). Applying this rule, we get the equation to be: \(f'(t) = (t^{3 / 2})' \log _{2} \sqrt{t+1} + t^{3 / 2} (\log _{2} \sqrt{t+1})'\).
03

Compute the Derivatives for Each Part

(a) For the first part, take the derivative of \(t^{3 / 2}\) to obtain \(\frac{3}{2}t^{1 / 2}\). (b) For the second part, the derivative of \(\log _{2} \sqrt{t+1}\) involves chain rule. The derivative of logarithm function \(\log_2 a\) is \(\frac{1}{\ln2}\cdot \frac{1}{a}\). Apply this rule along with the derivative of \(\sqrt{t+1}\) that is \(\frac{1}{2\sqrt{t+1}}\), we get \((\log _{2} \sqrt{t+1})' = \frac{1}{\ln2}\cdot \frac{1}{\sqrt{t+1}} \cdot \frac{1}{2\sqrt{t+1}} = \frac{1}{2\ln2(t+1)}\). Substituting these into the equation from Step 2.
04

Put It All Together

Substitute the computed derivatives from Step 3 into the equation from Step 2 to obtain the final derivative: \(f'(t) = \frac{3}{2}t^{1 / 2} \log _{2} \sqrt{t+1} + \frac{t^{3 / 2}}{2\ln2(t+1)}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
The product rule is a fundamental concept used to find the derivative of a product of two functions. Imagine you have two differentiable functions, say \( f(t) \) and \( g(t) \). If these functions are multiplied together, their product, denoted by \( F(t) = f(t)g(t) \), requires a special rule to differentiate. The product rule comes to the rescue and is defined as:
  • If \( F(t) = f(t)g(t) \), then the derivative \( F'(t) \) is given by \( F'(t) = f'(t)g(t) + f(t)g'(t) \).
This means, to find the derivative of two multiplied functions: - Differentiate the first function while keeping the second fixed. - Then, keep the first function fixed and differentiate the second. - Finally, sum the results from both steps.
Chain Rule
The chain rule is essential when dealing with composite functions, which are functions within functions. Suppose you have a function \( h(t) \) that can be expressed as \( h(t) = f(g(t)) \). This type involves a composition, where \( g(t) \) is inside \( f(t) \). For the derivative, the chain rule tells us to:
  • First, differentiate the outer function \( f \) with respect to \( g \), keeping \( g(t) \) as it is.
  • Then, multiply by the derivative of the inner function \( g(t) \).
It is effectively a way of "chaining" together the derivatives of nested functions. In the example of differentiating \( \log_2 \sqrt{t+1} \), the chain rule is used to address both the logarithmic part and the square root part.
Logarithmic Differentiation
Logarithmic differentiation is a technique often used for differentiating functions involving logarithms, like \( \log_2 \sqrt{t+1} \). This method can simplify complex differentiations:
  • First, express the given logarithm in a base that is easier to handle, such as the natural logarithm, ln.
  • Use the fact that for any function \( a \), the derivative of \( \log_2 a \) is \( \frac{1}{\ln(2)} \cdot \frac{1}{a} \).
In our example, identifying the expression inside the logarithm \( \sqrt{t+1} \) and applying logarithmic rules simplifies its differentiation, making it straightforward to apply the chain rule subsequently. By converting to natural logarithms, you can leverage known derivatives to facilitate the computation.
Function Composition
Function composition involves combining two or more functions such that the output of one function becomes the input of another. Mathematically, if you have two functions, \( f(t) \) and \( g(t) \), the composition is expressed as \( f(g(t)) \).
  • This concept is crucial when dealing with multi-layered expressions, as it allows breaking them down step-by-step.
  • The composite function \( f(g(t)) \) requires careful application of the chain rule to determine its derivative. It鈥檚 like a nesting doll, opening one layer at a time.
In the context of our differentiation problem, the function \( \log_2 \sqrt{t+1} \) is a perfect example of function composition, as it consists of a logarithmic outer layer and a square root inner layer.

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Most popular questions from this chapter

Find any relative extrema of the function. Use a graphing utility to confirm your result. $$ f(x)=x \sinh (x-1)-\cosh (x-1) $$

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Linear and Quadratic Approximations Use a computer algebra system to find the linear approximation \(P_{1}(x)=f(a)+f^{\prime}(a)(x-a)\) and the quadratic approximation \(P_{2}(x)=f(a)+f^{\prime}(a)(x-a)+\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}\) of the function \(f\) at \(x=a .\) Use a graphing utility to graph the function and its linear and quadratic approximations. $$ f(x)=\cosh x, \quad a=0 $$

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).
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