Problem 29 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ x^{2} \operatorname{arcsec}(x) $$

Step-by-Step Solution

Answer

The derivative of \(x^2\operatorname{arcsec}(x)\) is \(2x \operatorname{arcsec}(x) + \frac{x}{\sqrt{x^2 - 1}}\)."

Step 1: Identify the Rule to Use

To differentiate the expression, we will use the product rule. The product rule states that if you have two functions multiplied together, such as \(u(x) v(x)\), the derivative is \(u'(x) v(x) + u(x) v'(x)\). Here, \(u(x) = x^2\) and \(v(x) = \operatorname{arcsec}(x)\).

Step 2: Differentiate the First Function

Differentiate \(u(x) = x^2\) with respect to \(x\). Using the power rule, the derivative of \(x^2\) is \(2x\). Thus, \(u'(x) = 2x\).

Step 3: Differentiate the Second Function

To find the derivative of \(v(x) = \operatorname{arcsec}(x)\), recall the derivative formula: \(\frac{d}{dx}[\operatorname{arcsec}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}}\) for \(|x| > 1\). Thus, \(v'(x) = \frac{1}{|x|\sqrt{x^2 - 1}}\).

Step 4: Apply the Product Rule

Using the product rule, \(u'(x)v(x) + u(x)v'(x)\), substitute the derivatives and the original functions:\(u'(x) = 2x\), \(v(x) = \operatorname{arcsec}(x)\), \(u(x) = x^2\), \(v'(x) = \frac{1}{|x|\sqrt{x^2 - 1}}\).The derivative is:\[ 2x \cdot \operatorname{arcsec}(x) + x^2 \cdot \frac{1}{|x|\sqrt{x^2 - 1}} \]

Step 5: Simplify the Expression

Now, simplify the expression from Step 4:First term remains as: \(2x \cdot \operatorname{arcsec}(x)\).Second term simplifies as follows: \(\frac{x^2}{|x|\sqrt{x^2 - 1}} = \frac{x}{\sqrt{x^2 - 1}}\) since \(\frac{x^2}{|x|} = x\) for \(x > 1\).Finally, the derivative simplifies to:\[2x \operatorname{arcsec}(x) + \frac{x}{\sqrt{x^2 - 1}}\]

Key Concepts

Product Rule in Differentiation

When dealing with differentiating a product of two functions, the product rule is your best friend. This essential calculus tool makes it easy to handle such expressions.

The product rule states that if we have two functions, say \( u(x) \) and \( v(x) \), and they are multiplied together, their derivative can be found by:

• Differentiating \( u(x) \) to get \( u'(x) \)
• Multiplying \( u'(x) \) by the original \( v(x) \)
• Differentiating \( v(x) \) to get \( v'(x) \)
• Multiplying \( v'(x) \) by the original \( u(x) \)

Adding these two results gives you the derivative of the function pair: \[ u'(x) v(x) + u(x) v'(x)\]This rule is simple yet powerful, allowing for complex expressions to be differentiated with ease.

Arcsecant Differentiation

The arcsecant function, denoted as \( \operatorname{arcsec}(x) \), can be a tricky one to differentiate because it's not as commonly encountered as other trigonometric inverses. However, once you know the differentiation rule, it becomes straightforward.

The derivative of \( \operatorname{arcsec}(x) \) is given by:\[\frac{d}{dx}[\operatorname{arcsec}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}}\]This formula works for \(|x| > 1\), which is the domain restriction for the arcsecant function.

To apply this in practical problems, ensure that you substitute the function correctly within its domain. The formula involves both the absolute value of \( x \) and the expression \( \sqrt{x^2 - 1} \), reflecting the nature of the function itself.

Power Rule Simplified

The power rule is one of the first differentiation rules anyone learns. It's simple yet crucial for handling polynomial expressions.

The rule states that for any function \( x^n \), where \( n \) is a constant, the derivative is:\[\frac{d}{dx}[x^n] = nx^{n-1}\]For example, if \( u(x) = x^2 \), applying the power rule tells us that the derivative \( u'(x) \) is \(2x\).

Remember:

• Subtract \( 1 \) from the exponent.
• Multiply the result by the original exponent \( n \).

This rule makes differentiating terms like \( x^3 \), \( x^5 \), or even \( x^{100} \) an absolute breeze!