Question

$$ \text { } , \text { find the indicated derivative. } $$ $$ D_{\theta} \sqrt{\log _{10}\left(3^{\theta^{2}-\theta}\right)} $$

Short answer

The derivative is \( \frac{(2\theta - 1) \log_{10}(3)}{2\sqrt{(\theta^2 - \theta) \log_{10}(3)}} \)."

Step-by-step solution

Understanding the Problem

We are asked to find the derivative of the function \( \sqrt{\log_{10}(3^{\theta^2 - \theta})} \) with respect to \( \theta \). This involves a combination of chain rule, power rule, and logarithmic differentiation.

Simplifying the Expression

Simplify the expression inside the square root first. We know from logarithmic identities that \( \log_b(a^c) = c \log_b(a) \). Applying this, we get: \( \log_{10}(3^{\theta^2 - \theta}) = (\theta^2 - \theta) \log_{10}(3) \).

Differentiate the Simplified Expression

Now focus on differentiating \( \sqrt{x} \) where \( x = (\theta^2 - \theta) \log_{10}(3) \). Use the derivative of \( \sqrt{x} \), which is \( \frac{1}{2\sqrt{x}} \cdot \frac{dx}{d\theta} \).

Differentiate Inside the Derivative

Differentiate \( x = (\theta^2 - \theta) \log_{10}(3) \) with respect to \( \theta \). The derivative is \( \log_{10}(3) \cdot (2\theta - 1) \).

Complete the Derivative Calculation

Substitute \( dx/d\theta = \log_{10}(3) \cdot (2\theta - 1) \) into the derivative of \( \sqrt{x} \):\[ D_{\theta} \left( \sqrt{\log_{10}(3^{\theta^2 - \theta})} \right) = \frac{1}{2\sqrt{(\theta^2 - \theta) \log_{10}(3)}} \cdot \log_{10}(3) \cdot (2\theta - 1). \]

Key concepts

Chain Rule

When you have a complex function composed of several simpler functions, the chain rule is your friend. It helps you find the derivative of this composite function by multiplying the derivatives of the inner and outer functions. Let's break this down with a simple setup: if you have a function like \( f(g(x)) \), the chain rule tells you that the derivative, \( f'(g(x)) \), is \( f'(g(x)) \cdot g'(x) \). This multiplication allows you to "chain" the derivative of the outer function with the derivative of the inner function.

• First, identify the outer and inner functions. The outer function here is \( \sqrt{} \).
• The inner function is \( \log_{10}(3^{\theta^2 - \theta}) \).
• Differentiate the outer function: the derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).
• Then, multiply it with the derivative of the inner function.

Using the chain rule allows you to handle layered functions efficiently. Remember, it's like peeling layers from an onion—one layer at a time.

Logarithmic Differentiation

Logarithmic differentiation is a technique used when you're dealing with complex functions, especially those involving exponents and logs. In this exercise, it helped us handle the \( \log_{10}(3^{\theta^2 - \theta}) \) part. This method is handy because by taking natural logs, it often turns multiplication into addition, simplifying the process. It's like turning a tangled mess into a straightforward path.

• First, apply logarithmic identities: \( \log_b(a^c) = c \log_b(a) \). Here we have \( 3^{\theta^2 - \theta} \).
• Using the identity, simplify to \( (\theta^2 - \theta) \log_{10}(3) \).
• Take the derivative after simplifying, which becomes easier with this method.

This technique saves time and avoids errors by reducing the complexity of expressions with exponential forms.

Power Rule

The power rule is one of the simplest ways to find the derivative of a basic power function, but it's incredibly powerful (no pun intended) in calculus. Whenever you have a function like \( x^n \), the derivative, according to the power rule, is \( nx^{n-1} \). This is especially useful for polynomial functions. Think of it as a quick formula you can apply to get results fast.

• If you have \( \theta^2 - \theta \), treat each term independently.
• For \( \theta^2 \), apply the power rule to get \( 2\theta \).
• For \( -\theta \), treat it as \( -1\cdot\theta^1 \), leading to \( -1 \).

In this exercise, using the power rule allowed us to quickly find the derivative of the polynomial portion within the logarithmic differentiation step. It's like the calculus version of a Swiss Army knife: always ready and multifunctional.