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In calculus, the chain rule is a fundamental technique for finding derivatives of composite functions. The chain rule allows you to differentiate a "function within a function", often seen where there's an inner and an outer function connected.
For example, in the function \( y = (10x+1)^{\frac{1}{2}} \), the outer function is \( u^{\frac{1}{2}} \) and the inner function is \( u = 10x + 1 \). The chain rule tells us that the derivative is the product of the derivative of the outer function and the derivative of the inner function.
Let's break this down:

• The derivative of the outer function \( (u^{\frac{1}{2}})' \) is \( \frac{1}{2}u^{-\frac{1}{2}} \).
• The derivative of the inner function \( (10x+1)' \) is \( 10 \).

So, by applying the chain rule, we effectively multiply these two derivatives together to solve for the composite derivative. This step-by-step approach means initiating with the outer function, shifting to the inner, and ensuring each differentiation is executed correctly.

Exponential form refers to expressing numbers using an exponent, which simplifies calculations and analysis, especially in differentiation.
In our exercise, the original function \( y = \sqrt{10x+1} \) was transformed into \( y = (10x+1)^{\frac{1}{2}} \) to facilitate easier application of the chain rule. This transformation shows how roots, like square roots, can be re-expressed using exponents.
Here's the breakdown:

• Using the property \( \sqrt{a} = a^{\frac{1}{2}} \), any square root \( \sqrt{b} \) can be written as \( b^{\frac{1}{2}} \).
• This form not only simplifies the differentiation process but also aligns with standard calculus techniques for managing exponents.

By rewriting using exponential form, the process follows standard rules of differentiating power functions, thus avoiding potential confusion and streamlining calculations.

Simplification is about reducing expressions to their simplest form, often aiming to make them more interpretable and convenient for further computation or application.
After applying differentiation techniques to a function, the resulting derivative might be complex or cumbersome in its algebraic form, as with \( \frac{1}{2}(10x+1)^{-\frac{1}{2}} \cdot 10 \).
Here's how simplification plays a role in the exercise:

• Initially, multiplying gives \( \frac{10}{2} \), resulting in \( \frac{10}{2(10x+1)^{\frac{1}{2}}} \).
• Further simplification yields \( \frac{5}{\sqrt{10x+1}} \) as the most streamlined expression.

Such simplification not only makes the result visually cleaner but also eases subsequent steps in problem-solving or analyses. Always aim to maintain equivalency while converting complex expressions into their simplest forms, a crucial skill in math and calculus.