Problem 93 - Calculus and its applications

Question

Differentiate. $$ f(t)=\sqrt{3 t+\sqrt{t}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(t) = \frac{3}{2\sqrt{3t + \sqrt{t}}} + \frac{1}{4\sqrt{t} \cdot \sqrt{3t + \sqrt{t}}} \).

Step 1: Identify the Outer Function

Recognize that the function \( f(t) = \sqrt{3t + \sqrt{t}} \) is composed of an outer function \( u = \sqrt{u} \) and an inner function \( u = 3t + \sqrt{t} \). We will use the chain rule to differentiate it.

Step 2: Differentiate the Outer Function

Differentiate the outer function with respect to the inner function \( u \). The derivative of \( \sqrt{u} \) with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \).

Step 3: Differentiate the Inner Function

Differentiate the inner function \( 3t + \sqrt{t} \) with respect to \( t \). The derivative of \( 3t \) is 3. The derivative of \( \sqrt{t} \) is \( \frac{1}{2\sqrt{t}} \), so the derivative of the inner function is \( 3 + \frac{1}{2\sqrt{t}} \).

Step 4: Apply the Chain Rule

Apply the chain rule which states that the derivative of a composite function \( f(g(t)) \) is \( f'(g(t)) \cdot g'(t) \). Thus, the derivative of \( f(t) \) is \( \frac{1}{2\sqrt{3t + \sqrt{t}}} \cdot (3 + \frac{1}{2\sqrt{t}}) \).

Step 5: Simplify the Expression

Simplify the expression of the derivative: \[ f'(t) = \frac{3 + \frac{1}{2\sqrt{t}}}{2\sqrt{3t + \sqrt{t}}} = \frac{3}{2\sqrt{3t + \sqrt{t}}} + \frac{1}{4\sqrt{t} \cdot \sqrt{3t + \sqrt{t}}} \]. Therefore, the derivative of the function is this simplified expression.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus when differentiating composite functions. It’s crucial for finding the derivative of a function that is made up of other functions.
To say it simply, the chain rule is used when you have a function inside another function. The rule helps us differentiate such a setup systematically. It states that the derivative of a composite function \( f(g(t)) \) can be found by multiplying the derivative of the outer function \( f' \) evaluated at the inner function \( g(t) \), by the derivative of the inner function \( g'(t) \). This is written as:

• \( f'(g(t)) \cdot g'(t) \).

For example, in the function \( f(t) = \sqrt{3t + \sqrt{t}} \), the outer function is \( \sqrt{u} \) and the inner function is \( 3t + \sqrt{t} \).
To find the derivative, you first differentiate the outer function with respect to the inner function, and then the inner function with respect to \( t \). Applying the chain rule combines these derivatives effectively.

Composite Functions

Composite functions are an important concept to grasp for applying the chain rule. These are functions made by composing one function within another.
In easier terms, you take a function and plug it into another function. For example, \( f(t) = \sqrt{3t + \sqrt{t}} \) is a composite function where \( 3t + \sqrt{t} \) is substituted into \( \sqrt{u} \).
Composite functions are pivotal because differentiating them requires extra steps, usually involving the chain rule.

• The first step is to clearly identify both the outer function and the inner function.
• Then, differentiate each separately before applying the chain rule.

Recognizing composite functions often requires practice, but this skill helps in working out problems where a function is generated from the combination of different parts, as in this example.

Derivative of Square Root

When differentiating a square root function, you're dealing with exponents. The square root function \( \sqrt{u} \) can be rewritten as \( u^{1/2} \).
Using the basic rules of differentiation, the derivative of \( u^{1/2} \) with respect to \( u \) is \( \frac{1}{2}u^{-1/2} \), which simplifies to:

• \( \frac{1}{2\sqrt{u}} \).

This derivative form comes in handy when dealing with composite functions where a square root is the outer function, like in our example above with \( f(t) = \sqrt{3t + \sqrt{t}} \).
Knowing how to differentiate square roots is a small yet crucial piece of the puzzle. It enables applying the chain rule smoothly when dealing with composite functions, especially where square roots are involved. Once you master this concept, handling functions involving square roots becomes much more straightforward.