Problem 14 - Calculus

Question

Find the derivatives of the functions in \(1-26\). $$ \sqrt{x}(\sqrt{x}+1)(\sqrt{x}+2) $$

Step-by-Step Solution

Answer

The derivative is \(\frac{3}{2}\sqrt{x} + 3 + \frac{1}{\sqrt{x}}\).

Step 1: Expand the Expression

First, expand the given expression \(\sqrt{x}(\sqrt{x}+1)(\sqrt{x}+2)\). Begin by expanding the terms inside the parentheses. Multiply \((\sqrt{x}+1)(\sqrt{x}+2)\):\[(\sqrt{x} + 1)(\sqrt{x} + 2) = \sqrt{x} \cdot \sqrt{x} + \sqrt{x} \cdot 2 + 1 \cdot \sqrt{x} + 1 \cdot 2\]\[= x + 2\sqrt{x} + \sqrt{x} + 2\]\[= x + 3\sqrt{x} + 2\]

Step 2: Multiply by \(\sqrt{x}\)

Now multiply the expanded expression by \(\sqrt{x}\):\[\sqrt{x}(x + 3\sqrt{x} + 2)\]Distribute \(\sqrt{x}\) across each term:\[= \sqrt{x} \cdot x + \sqrt{x} \cdot 3\sqrt{x} + \sqrt{x} \cdot 2\]\[= x\sqrt{x} + 3x + 2\sqrt{x}\].

Step 3: Rewrite in a Form for Differentiation

Rewrite each term in the form of power functions to ease differentiation:\[= x^{3/2} + 3x + 2x^{1/2}\].

Step 4: Differentiate Each Term

Differentiate each term using the power rule \(\frac{d}{dx}[x^n] = nx^{n-1}\):For \(x^{3/2}\), the derivative is:\[\frac{d}{dx}[x^{3/2}] = \frac{3}{2}x^{1/2}\]For \(3x\), the derivative is:\[\frac{d}{dx}[3x] = 3\]For \(2x^{1/2}\), the derivative is:\[\frac{d}{dx}[2x^{1/2}] = 2 \cdot \frac{1}{2}x^{-1/2} = x^{-1/2}\].

Step 5: Combine Derivatives

Combine the derivatives to form the derivative of the original function:\[= \frac{3}{2}x^{1/2} + 3 + x^{-1/2}\].

Step 6: Simplify the Derivative

Leave the derivative in terms of radicals or simplify further if required:\[= \frac{3}{2}\sqrt{x} + 3 + \frac{1}{\sqrt{x}}\].

Key Concepts

power rule

The power rule is a fundamental tool in differentiation, which simplifies finding the derivative of a power function. It is applicable to any function of the form \(x^n\), where \(n\) is any real number. The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\). This means you multiply the exponent \(n\) by the coefficient and then reduce the exponent by one.
For example, if you have a function \(x^{3/2}\), applying the power rule will give you the derivative \(\frac{3}{2}x^{1/2}\).
Every time you differentiate a term in the form \(x^n\), you can apply this straightforward rule, making differentiation of power functions quick and efficient.

differentiation

Differentiation is the mathematical process of finding a derivative of a function. The derivative represents the rate at which a function changes at any given point.

• It provides significant insights in various fields, such as physics for calculating velocity and acceleration.
• In simpler terms, if you think of a function as a curve, the derivative tells you how steep that curve is at any point.

To differentiate a complex function, like the one provided, the function first must be simplified (as shown in earlier steps), usually by expanding and rewriting in forms suitable for differentiation, such as using power functions.

radicals

Radicals involve expressions containing a square root, cube root, or any higher root. The symbol \(\sqrt{}\) or more generally \(\sqrt[n]{x}\) indicates a radical. These expressions can sometimes be tricky to handle in calculus.
A common first step with radicals is to transform them into power functions because differentiation rules, like the power rule, are better suited for power functions.
For example, \(\sqrt{x}\) can be rewritten as \(x^{1/2}\). This conversion simplifies differentiating expressions that include radicals, allowing you to apply the power rule efficiently.

expansion of expressions

Expanding expressions is an important step in simplifying functions before differentiation. By expanding, you break down complex expressions into simpler components.

• Start by applying distributive property: multiply everything inside one set of parentheses by everything in another.
• For instance, to expand \((\sqrt{x} + 1)(\sqrt{x} + 2)\), multiply each term in the first bracket by each term in the second bracket.

Through expansion, complicated expressions can be rewritten as sums of simpler terms, making the differentiation process much more manageable. Properly expanding the expression is crucial for correctly applying differentiation techniques later.