Video Transcript
Find d𝑦 by d𝑥, given that 𝑦
equals four times the natural log of 𝑥 plus three over four times the natural log
of 𝑥 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 d𝑦 by d𝑥. This says that the derivative of
the quotient of two differentiable functions 𝑢 and 𝑣 is 𝑣 times d𝑢 by d𝑥 minus
𝑢 times d𝑣 by d𝑥 all over 𝑣 squared. We let 𝑢 be equal to four times
the natural log of 𝑥 plus three. And 𝑣 is equal to the denominator
of our fraction. That’s four times the natural log
of 𝑥 minus seven. We then quote the general result
for the derivative of the natural log of 𝑥; it’s one over 𝑥. And since the derivative of a
constant is zero, we see that d𝑢 by d𝑥 is equal to four lots of one over 𝑥, which
is simply four over 𝑥. And similarly d𝑣 by d𝑥 is also
four over 𝑥. d𝑦 by d𝑥 is equal to 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥
all over 𝑣 squared.
Let’s multiply the numerator and
denominator of this fraction by 𝑥 to simplify. When we do, we see that d𝑦 by d𝑥
is equal to four times four times the natural log of 𝑥 minus seven minus four times
four times the natural log of 𝑥 plus three all over 𝑥 times four times the natural
log of 𝑥 minus seven squared. We distribute the parentheses on
our numerator. And we see that we have 16 times
the natural log of 𝑥 minus 16 times the natural log of 𝑥 which gives us zero. And we found the derivative of our
quotient. It’s negative 40 over 𝑥 times four
times the natural log of 𝑥 minus seven squared.
