Question Video: Differentiating Rational Functions at a Point Using the Quotient Rule | Nagwa Question Video: Differentiating Rational Functions at a Point Using the Quotient Rule | Nagwa

# Question Video: Differentiating Rational Functions at a Point Using the Quotient Rule Mathematics • Second Year of Secondary School

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If the function π(π₯) = (5π₯ + 7)/(4π₯ + 2), determine its rate of change when π₯ = 2.

02:07

### Video Transcript

If the function π of π₯ equals five π₯ plus seven over four π₯ plus two, determine its rate of change when π₯ is equal to two.

We recall the definition for the rate of change of the function or its derivative. Itβs the limit as β approaches zero of π of π plus β minus π of π over β assuming that limit exists. In this question, π of π₯ is equal to five π₯ plus seven over four π₯ plus two. And we want to find the rate of change when π₯ is equal to two. So weβre going to let π be equal to two. So we need to evaluate π prime of two, the rate of change or the derivative of our function, when π₯ equals two.

By our definition, itβs the limit as β approaches zero of π of two plus β minus π of two over β. Letβs work out π of two plus β and π of two. To find π of two plus β, we replace each instance of π₯ in our original function with two plus β. And we get five times two plus β plus seven over four times two plus β plus two. And when we distribute our parentheses and simplify, we get 17 plus five β over 10 plus four β. Similarly, π of two is five times two plus seven over four times two plus two which is 17 tenths. And we now see that π prime of two is the limit as β approaches zero of the difference between these all over β.

There are two fractions in our numerator. So weβre going to simplify by creating a common denominator there. Weβll multiply the numerator and denominator of the first fraction on the numerator by 10 and the second fraction on our numerator by 10 plus four β. And when we do, we achieve the numerator shown. Well, this doesnβt make a lot of sense. But actually, dividing this entire fraction by β is the same as timesing it by one over β. So we rewrite our denominator as a β times 100 plus 40β. And then, we simplify a numerator to negative 18β. And you might now see that we can simplify further by dividing through by β.

And weβre now ready to perform direct substitution. By replacing β with zero, we find that π prime of two is negative 18 over 100, which simplifies to negative nine over 50. The rate of change of our function π of π₯ when π₯ is equal to two is negative nine over 50.

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