Problem 47 Differentiate each function. $... [FREE SOLUTION] (2024)

Short Answer

The Quotient Rule for Differentiation

Computing Derivatives of Numerator and Denominator

Before we can use the Quotient Rule, we need to find the derivatives of the numerator and the denominator separately. Let's break it down:
Consider the function \( g(t) = \frac{-t^{2}+3t+5}{t^{2}-2t+4} \). Here, the numerator is \( f(t) = -t^2 + 3t + 5 \) and the denominator is \( g(t) = t^2 - 2t + 4 \).
First, find the derivative of the numerator (\( f(t) \)):
\[ f'(t) = \frac{d}{dt}(-t^2 + 3t + 5) = -2t + 3\] Next, find the derivative of the denominator (\( g(t) \)):
\[ g'(t) = \frac{d}{dt}(t^2 - 2t + 4) = 2t - 2 \]
Now we've got \( f'(t) = -2t + 3 \) and \( g'(t) = 2t - 2 \). These are the components we need for the Quotient Rule. Always double-check your differentiation steps to ensure you’ve done them correctly.

Simplifying the Derived Function

After applying the Quotient Rule, we get a new function that might look complex. The next step is to simplify it as much as possible. Let's see this in action with our function:
Substitute \( f(t) \), \( f'(t) \), \( g(t) \), and \( g'(t) \) into the Quotient Rule:
\[ g'(t) = \frac{(-2t + 3)(t^2 - 2t + 4) - (-t^2 + 3t + 5)(2t - 2)}{(t^2 - 2t + 4)^2} \]Next, expand and simplify the numerator:
\[ g'(t) = \frac{(-2t^3 + 4t^2 - 8t + 3t^2 - 6t + 12) - (-2t^3 + 2t^2 + 6t^2 - 6t + 10t - 10)}{(t^2 - 2t + 4)^2} \]Combine like terms:
\[ g'(t) = \frac{-2t^3 + 7t^2 - 14t + 12 + 2t^3 - 8t^2 + 10}{(t^2 - 2t + 4)^2} \]
Further simplification yields:
\[ g'(t) = \frac{-t^2 - 14t + 22}{(t^2 - 2t + 4)^2} \]Simplifying makes it easier to understand the behavior of the derivative function. It’s like tidying up after cooking – it helps you see what you’ve created more clearly!