### Video Transcript

Evaluate the rate of change of π of π₯ equals six π₯ squared plus seven all over seven π₯ at π₯ equals three.

Remember the definition for the instantaneous rate of change of a function at a point π, also known as its derivative at that point, is given by the limit as β approaches zero of π of π plus β minus π of π all over β. In this question, π of π₯ is equal to six π₯ squared plus seven all over seven π₯. And weβre looking to find the instantaneous rate of change when π₯ is equal to three. So weβre going to let π be equal to three. We can replace π with three. And we see that the rate of change of the function at π₯ equals three is given by the limit as β approaches zero of π of three plus β minus π of three all over β.

Weβre now going to substitute three plus β and three into our function. π of three plus β is then six lots of three plus β squared plus seven over seven lots of three plus β. And then, π of three is six times three squared plus seven over seven times three. Weβre going to distribute the parentheses in the numerator of the first fraction, and weβre going to evaluate the numerator of the second. When we do, we see that π prime of three, the derivative of our function and therefore the rate of change π₯ equals three, is equal to the limit as β approaches zero of six β squared plus 36β plus 61 over seven times three plus β minus 61 over seven times three all over β.

Now, I specifically didnβt evaluate the denominator of these two fractions. And that was to remind me that each denominator had a factor of seven. So when I subtract by creating a common denominator, I simply need to multiply the numerator and denominator of my first fraction by three and of my second fraction by three plus β. That gives me a common denominator of 21 times three plus β. And then, the numerator is 18β squared plus 108β plus 183 minus 183 minus 61β.

Now, since dividing by β is the same as multiplying by one over β, we can rewrite our denominator as 21β times three plus β. We then spot that 183 minus 183 is zero. And we also see that we can divide both the numerator and the denominator of our fraction by β. That leaves us with the limit as β approaches zero of 18β plus 108 minus 61 over 21 times three plus β. And the numerator here further simplifies to 18β plus 47. At this stage, weβre now ready to use direct substitution to evaluate our limit. Replacing β with zero and we find that π prime of three is 18 times zero plus 47. Thatβs just 47 over 21 times three plus zero. We evaluate the denominator as 63.

And we found the rate of change of our function at π₯ equals three to be 47 over 63.