Problem 36 - Calculus Early Transcendentals

Question

Derivatives Evaluate the derivatives of the following functions. \(h(t)=(\sin t)^{\sqrt{t}}\)

Step-by-Step Solution

Answer

Question: Find the derivative of the function \(h(t)=(\sin t)^{\sqrt{t}}\). Answer: The derivative of the function \(h(t)=(\sin t)^{\sqrt{t}}\) is \(\frac{dh}{dt}=(\sin t)^{\sqrt{t}}(\sqrt{t}\cdot\frac{\cos t}{\sin t}+\ln(\sin t)\cdot\frac{1}{2\sqrt{t}})\).

Step 1: Rewrite the expression

We can rewrite the given function as: \(h(t)=e^{\sqrt{t}\ln(\sin t)}\)

Step 2: Differentiate using the chain rule

We need to find the derivative of the function \(h(t)\) with respect to \(t\) using the chain rule: \(\frac{dh}{dt}=\frac{d(e^{\sqrt{t}\ln(\sin t)})}{dt}=e^{\sqrt{t}\ln(\sin t)}\cdot\frac{d(\sqrt{t}\ln(\sin t))}{dt}\) Now, we need to find the derivative of the exponent, \(\sqrt{t}\ln(\sin t)\), with respect to \(t\).

Step 3: Differentiate the exponent term

Using the product rule, find the derivative of the exponent term: \(\frac{d(\sqrt{t}\ln(\sin t))}{dt}=\sqrt{t}\cdot\frac{d(\ln(\sin t))}{dt}+\ln(\sin t)\cdot\frac{d(\sqrt{t}))}{dt}\) Now we need to find the derivatives of \(\ln(\sin t)\) and \(\sqrt{t}\).

Step 4: Differentiate \(\ln(\sin t)\) and \(\sqrt{t}\)

Using the chain rule, differentiate \(\ln(\sin t)\): \(\frac{d(\ln(\sin t))}{dt}=\frac{1}{\sin t}\cdot\frac{d(\sin t)}{dt}=\frac{\cos t}{\sin t}\) Differentiate \(\sqrt{t}\) using the power rule: \(\frac{d(\sqrt{t})}{dt}=\frac{1}{2\sqrt{t}}\)

Step 5: Substitute the derivatives into the exponent term

Substitute the derivative terms from Step 4 back into the expression derived in Step 3: \(\frac{d(\sqrt{t}\ln(\sin t))}{dt}=\sqrt{t}\cdot\frac{\cos t}{\sin t}+\ln(\sin t)\cdot\frac{1}{2\sqrt{t}}\)

Step 6: Substitute exponent term back into derivative

Now, substitute this expression back into the derivative expression obtained in Step 2: \(\frac{dh}{dt}=e^{\sqrt{t}\ln(\sin t)}\cdot(\sqrt{t}\cdot\frac{\cos t}{\sin t}+\ln(\sin t)\cdot\frac{1}{2\sqrt{t}})\)

Step 7: Simplify the final expression

Finally, substitute \(h(t)=(\sin t)^{\sqrt{t}}\) back into the expression: \(\frac{dh}{dt}=(\sin t)^{\sqrt{t}}(\sqrt{t}\cdot\frac{\cos t}{\sin t}+\ln(\sin t)\cdot\frac{1}{2\sqrt{t}})\) This is the derivative of the function \(h(t)=(\sin t)^{\sqrt{t}}\).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. Imagine you're driving a car: while your speedometer shows your speed at each instant, the derivative of your distance over time provides the mathematical equivalent of your speed. In mathematics, derivatives provide us with essential tools for analyzing and understanding dynamic systems. They let us calculate things like

• the rate of change,
• identification of maximum and minimum points in functions, and
• understanding the slope of curves at any given point.

In our exercise, we are tasked with finding the derivative of the function \( h(t) = (\sin t)^{\sqrt{t}} \). This involves not only basic differentiation but also more advanced techniques like the chain rule and the product rule. Mastering derivatives means being able to deconstruct complex expressions into simpler ones, often requiring multiple layers of differentiation techniques.

The Product Rule Explained

The product rule is an essential technique in differential calculus used when finding the derivative of a product of two functions. It can be simply stated as: if you have two functions \( u(t) \) and \( v(t) \), their derivative when multiplied together is given by:
\[ \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \]This rule is particularly useful because, in real-life functions, we often encounter products of separate expressions. In our exercise, when differentiating the exponent \( \sqrt{t}\ln(\sin t) \), the product rule is employed. Let's break it down:

• \( \sqrt{t} \) is one function, while \( \ln(\sin t) \) is another.
• The product rule tells us to take the derivative of each function separately and then combine them according to the formula above.

This allows us to efficiently manage the complexity of these functions, hence simplifying the differentiation process. Good practice with the product rule complements other techniques, such as the chain rule, in dealing with complex functions.

Differentiation Techniques

Differentiation techniques provide us with methods to tackle a variety of problems in calculus effectively. These include various rules like power, product, quotient, and chain rules, each serving different kinds of functions. In the given exercise, several differentiation techniques are employed to find the derivative of \( h(t)=(\sin t)^{\sqrt{t}} \). Here’s how they play a role:

• Chain Rule: This powerful method allows us to differentiate a function that is composed of nested functions, such as \( e^{\sqrt{t}\ln(\sin t)} \). Using the chain rule, we differentiate the outer function separately from the inner one.
• Product Rule: As mentioned earlier, this handles multiplication between two functions, particularly in the \( \sqrt{t}\ln(\sin t) \) part of our derivative.
• Power Rule: This is used to differentiate \( \sqrt{t} \) by rewriting it as \( t^{1/2} \) and applying the power rule: \( \frac{d}{dt}[t^n] = nt^{n-1} \).

Combining these techniques allows us to systematically take apart complex functions and understand their instantaneous behavior. Each technique is like a tool in a toolbox, and knowing when and how to apply them is key to mastering calculus.