91Ó°ÊÓ

The ability to differentiate products of functions is vital in calculus. The product rule is the tool we use for this purpose.
When you have a function that is the product of two other functions, say \( u\) and \( v\), the product rule says that the derivative of \( uv \) is given by:
\[ (uv)' = u'v + uv' \. \]
In this problem, we applied the product rule to the term \( y_1 = (5x + 1)(x^2 - 1)\). Here, \( u = 5x + 1\) and \( v = x^2 - 1\). We then found their derivatives, which are \( u' = 5 \) and \( v' = 2x \), and applied the product rule formula:
\[ y_1' = (5)(x^2 - 1) + (5x + 1)(2x) \. \]
This gives us the intermediate result \(5x^2 - 5 + 10x^2 + 2x\). Simplifying combines similar terms to get: \( 15x^2 + 2x - 5 \).

Polynomial functions are among the simplest to differentiate.
For a polynomial \( ax^n \), the derivative is found using the power rule, which says:
\[ \frac{d}{dx}(ax^n) = anx^{n-1} \. \]
In the exercise, the second term of the function was \( y_2 = \frac{2x + 1}{3} \). This can be seen as a simple polynomial divided by a constant.
We first rewrite it as \( \frac{1}{3}(2x + 1) \), then differentiate term by term:
The derivative of \( 2x \) is 2, and the derivative of the constant 1 is 0.
Therefore, the overall derivative is: \[ \frac{1}{3}(2) = \frac{2}{3} \. \]This illustrates how the power rule and constant factor rule simplify the differentiation of polynomials.

Once you've found the derivatives of individual terms, the final step in differentiation problems like this is to combine them.
Let's summarize the derivatives we calculated:

• First term: \( 15x^2 + 2x - 5 \)
• Second term: \( \frac{2}{3} \)

We then add these together to get the final result for the derivative of the entire function:
\[ y' = 15x^2 + 2x - 5 + \frac{2}{3} \]
Combining all terms, we get:
\[ y' = 15x^2 + 2x - \frac{13}{3} \. \]
This step-by-step combination ensures all individual derivatives are correctly accounted for, giving us a clear final expression.