Problem 36 - Calculus and its applications

Question

Differentiate. $$ g(x)=\sqrt{x^{3}-x}\left(\log _{5} x\right) $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = \frac{(3x^2-1)\log_5(x)}{2\sqrt{x^3-x}} + \frac{\sqrt{x^3-x}}{x\ln(5)} \).

Step 1: Apply the Product Rule

To differentiate the function \( g(x) = \sqrt{x^3 - x} \cdot \log_5(x) \), recognize it as a product of two functions: \( u(x) = \sqrt{x^3 - x} \) and \( v(x) = \log_5(x) \). The product rule states \( (uv)' = u'v + uv' \).

Step 2: Differentiate \( u(x) = \sqrt{x^3 - x} \)

Rewrite \( u(x) \) as \( (x^3 - x)^{1/2} \). Use the chain rule for differentiation: \( u'(x) = \frac{1}{2}(x^3-x)^{-1/2}(3x^2-1) \) because the derivative of \( x^3 - x \) is \( 3x^2 - 1 \).

Step 3: Differentiate \( v(x) = \log_5(x) \)

Use the change of base formula for logarithms: \( \log_5(x) = \frac{\ln(x)}{\ln(5)} \). Differentiate to get \( v'(x) = \frac{1}{x \ln(5)} \).

Step 4: Combine Using the Product Rule

Substitute \( u'(x) \) and \( v'(x) \) into the product rule formula. The derivative is \( g'(x) = \left(\frac{1}{2}(x^3-x)^{-1/2}(3x^2-1)\right)\log_5(x) + \sqrt{x^3-x}\left(\frac{1}{x\ln(5)}\right) \).

Step 5: Simplify the Expression

Simplify \( g'(x) \) from Step 4 to make it more manageable, keeping separate terms: \( g'(x) = \frac{(3x^2-1)\log_5(x)}{2\sqrt{x^3-x}} + \frac{\sqrt{x^3-x}}{x\ln(5)} \).

Key Concepts

Product Rule

When you have a function that is the product of two separate functions, the derivative can't just be the product of their derivatives. Instead, you'll apply the Product Rule. This rule is essential in differentiation scenarios where tasks involve multiplying two or more functions together. To better understand, consider two functions, say \( u(x) \) and \( v(x) \). According to the Product Rule, the derivative of their product is given by :- Find the derivative of the first function \( u'(x) \) and multiply it by the second function \( v(x) \).- Then, find the derivative of the second function \( v'(x) \) and multiply by the first function \( u(x) \). - Add these results: \((uv)' = u'v + uv'\).In simpler terms, the Product Rule says to "derive one, keep the other, then switch." This method ensures all factors within the product are accounted for during differentiation. In our example, where \( g(x)=\sqrt{x^{3}-x}\log_{5}x \), you identify \( u(x) \) and \( v(x) \) as the two parts and proceed to differentiate each separately before combining the results to get \( g'(x) \).

Chain Rule

For functions nested within other functions, you use the Chain Rule to simplify the differentiation process. It’s like peeling an onion, where each layer can be thought of as another function.Once established, the outer function can be named \( f(x) \), while the inner function could be \( g(x) \). When combined, it's referred to as \( f(g(x)) \). The rule tells us to take the derivative of the outer function while keeping the inner function inside, then multiply by the derivative of the inner function. For instance, with \( u(x) = \sqrt{x^3-x} \), we can rewrite it as \( (x^3-x)^{1/2} \) and then apply the Chain Rule.- Differentiate the outer function—which is \( (x^3 - x)^{1/2} \)—as \( \frac{1}{2}(x^3 - x)^{-1/2} \).- Then differentiate the inside \( x^3 - x \) getting \( 3x^2 - 1 \).Multiply these together as \( u'(x) = \frac{1}{2}(x^3-x)^{-1/2}(3x^2-1) \), creating the link between the inner and outer functions' derivatives.

Logarithmic Differentiation

Logarithmic differentiation is a handy technique, especially when dealing with functions that involve logarithms, like our \( v(x) = \log_5(x) \). This approach is particularly useful when you want to differentiate a complicated function, where direct application of derivatives might be cumbersome.Start by recalling that any logarithm can be rewritten using the change of base formula. For instance, logs in another base such as \( \log_5(x) \) can be expressed as \( \frac{\ln(x)}{\ln(5)} \). By doing this, the differentiation becomes clearer and easily manageable.With this rewriting, the derivative follows the standard logarithmic rules. Knowing that the derivative of \( \ln(x) \) is \( \frac{1}{x} \), we apply that here:- Differentiate \( \ln(x) \) to get \( \frac{1}{x} \).- Because of the constant \( \ln(5) \) in the denominator, the result is \( v'(x) = \frac{1}{x \ln(5)} \).This method allows more streamlined calculus processes with integrals and derivatives, especially when you're multiplying or dividing functions, as seen in our example. By using the properties of logarithms, the differentiation process often becomes simpler.