Problem 11 - Thomas Calculus

Question

In Exercises \(7-12,\) find the indicated derivatives. $$ \frac{d p}{d q} \quad \text { if } \quad p=\frac{1}{\sqrt{q+1}} $$

Step-by-Step Solution

Answer

The derivative \( \frac{dp}{dq} \) is \(-\frac{1}{2(q+1)^{\frac{3}{2}}}\).

Step 1: Identify the Function

The given function is \( p = \frac{1}{\sqrt{q+1}} \). We need to find the derivative of \( p \) with respect to \( q \), i.e., \( \frac{dp}{dq} \).

Step 2: Rewrite the Function Using Exponents

Rewrite \( p = \frac{1}{\sqrt{q+1}} \) as \( p = (q+1)^{-\frac{1}{2}} \) using the property that \( \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}} \). This will make it easier to differentiate.

Step 3: Apply the Power Rule for Derivatives

Use the power rule \( \frac{d}{dx}[x^n] = n \cdot x^{n-1} \) to differentiate \( p = (q+1)^{-\frac{1}{2}} \). Applying this rule, we get:\[ \frac{dp}{dq} = -\frac{1}{2} \cdot (q+1)^{-\frac{1}{2}-1} \]

Step 4: Simplify the Expression

Simplify the expression from Step 3:\[ \frac{dp}{dq} = -\frac{1}{2} \cdot (q+1)^{-\frac{3}{2}} \]

Step 5: Express the Final Derivative in Original Terms

Rewrite the derivative back in radical form to match the original terms, if necessary. The simplified derivative is:\[ \frac{dp}{dq} = -\frac{1}{2} \cdot \frac{1}{(q+1)^{\frac{3}{2}}} \] which can be written as \(-\frac{1}{2(q+1)^{\frac{3}{2}}}\).

Key Concepts

Power Rule

When calculating derivatives, the power rule is an essential tool that makes differentiation easier, especially for polynomial functions. It states that for any function where the variable is raised to a power, such as \( x^n \), the derivative is \( n \cdot x^{n-1} \). This means you multiply by the exponent and then reduce the exponent by one.To apply the power rule effectively:

• Identify the exponent \( n \) in the function.
• Multiply the entire function by this exponent.
• Reduce the exponent by one to find the new power of the variable.

This simple yet powerful rule greatly simplifies the differentiation process and is widely used in calculus to find derivatives quickly. Its efficiency is evident in even more complex functions that can be rewritten to expose each term in the form needed for the power rule.

Exponent Rules

Exponent rules help in rearranging expressions into a form that's ready for differentiation. These rules dictate how powers of numbers or variables behave during multiplication, division, and raising to powers.Consider the function \( p = \frac{1}{\sqrt{q+1}} \). By using exponent rules, this can be rewritten as \( (q+1)^{-\frac{1}{2}} \). This transformation is vital, as it translates the original expression into an easier format for differentiation.Key exponent rules:

• \( x^{-a} = \frac{1}{x^a} \) - A negative exponent indicates a reciprocal.
• \( \sqrt{x} = x^{\frac{1}{2}} \) - A square root can be rewritten as a power of \( \frac{1}{2} \).
• \( (x^a)^b = x^{a\cdot b} \) - Powers of powers multiply exponents.

Using these rules simplifies complex expressions and makes the application of the power rule straightforward in calculus.

Differentiation Techniques

To master derivatives in calculus, understanding various differentiation techniques is crucial. These techniques include the basic rules such as the power rule, product rule, and quotient rule, among others. However, the starting point often includes simplifying expressions to make these techniques applicable.For the function \( p = \frac{1}{\sqrt{q+1}} \), the method began by rewriting the expression using the power rule and exponent laws to transform it into \( (q+1)^{-\frac{1}{2}} \). Once in this format, applying the power rule becomes straightforward, resulting in:\[ \frac{dp}{dq} = -\frac{1}{2} \cdot (q+1)^{-rac{3}{2}} \]Steps to effective differentiation:

• Rewrite Expressions: Use exponent rules to simplify.
• Apply the Power Rule: Use the rule for terms where the variable has a power.
• Further Techniques: Sometimes, employ product, quotient, or chain rules.

Grasping these techniques enables solving a wide array of differentiation problems efficiently and is a fundamental skill in calculus.