Video Transcript
Find dπ¦ by dπ₯, given that π¦
equals four times the natural log of π₯ plus three over four times the natural log
of π₯ minus seven.
In this question, we have a
fraction or a quotient. This tells us we can use the
quotient rule to find the derivative dπ¦ by dπ₯. This says that the derivative of
the quotient of two differentiable functions π’ and π£ is π£ times dπ’ by dπ₯ minus
π’ times dπ£ by dπ₯ all over π£ squared. We let π’ be equal to four times
the natural log of π₯ plus three. And π£ is equal to the denominator
of our fraction. Thatβs four times the natural log
of π₯ minus seven. We then quote the general result
for the derivative of the natural log of π₯; itβs one over π₯. And since the derivative of a
constant is zero, we see that dπ’ by dπ₯ is equal to four lots of one over π₯, which
is simply four over π₯. And similarly dπ£ by dπ₯ is also
four over π₯. dπ¦ by dπ₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯
all over π£ squared.
Letβs multiply the numerator and
denominator of this fraction by π₯ to simplify. When we do, we see that dπ¦ by dπ₯
is equal to four times four times the natural log of π₯ minus seven minus four times
four times the natural log of π₯ plus three all over π₯ times four times the natural
log of π₯ minus seven squared. We distribute the parentheses on
our numerator. And we see that we have 16 times
the natural log of π₯ minus 16 times the natural log of π₯ which gives us zero. And we found the derivative of our
quotient. Itβs negative 40 over π₯ times four
times the natural log of π₯ minus seven squared.
