Given that four 𝑥 squared minus
five 𝑥 plus eight 𝑦 squared minus three 𝑦 equals one, find 𝑑𝑦 by 𝑑𝑥.
We could try to rearrange this
equation so that we had 𝑦 in terms of 𝑥 and then differentiate in the normal
way. This is tricky because not only do
we have this minus three 𝑦 term, but also this eight 𝑦 squared. And so we have a quadratic in
𝑦. It’s possible to solve this
quadratic and find 𝑦 in terms of 𝑥 and then differentiate. But there is a better way.
As this equation is a relation
between the variables 𝑥 and 𝑦, we can differentiate both sides with respect to
𝑥. We can differentiate the left-hand
side term by term, using the fact that the derivative with respect to 𝑥 of 𝑎 times
𝑥 to the 𝑛 is 𝑎 times 𝑛 times 𝑥 to the 𝑛 minus one. We find that our first derivative
is eight 𝑥.
Similarly, 𝑑 by 𝑑𝑥 of five 𝑥 is
five. This term is slightly more
difficult to differentiate. How do you find 𝑑 by 𝑑𝑥 of a
function of 𝑦? Here, we can use the chain rule:
𝑑𝑓 by 𝑑𝑥 is 𝑑𝑓 by 𝑑𝑦 times 𝑑𝑦 by 𝑑𝑥. And so just using slightly
different notation here, we see that to differentiate a function of 𝑦 with respect
to 𝑥, we differentiate that function of 𝑦 with respect to 𝑦 and then we just
multiply by 𝑑𝑦 by 𝑑𝑥.
So what is 𝑑 by 𝑑𝑥 of eight 𝑦
squared? We differentiate eight 𝑦 squared
with respect to 𝑦 and get 16𝑦 and then you multiply by 𝑑𝑦 by 𝑑𝑥. Similarly, for 𝑑 by 𝑑𝑥 of three
𝑦, you differentiate three 𝑦 with respect to 𝑦 to get three and then multiply by
𝑑𝑦 by 𝑑𝑥. The right-hand side is more
straightforward. So the derivative of one with
respect to 𝑥 is just zero.
We now have an equation in terms of
𝑥, 𝑦, and 𝑑𝑦 by 𝑑𝑥 and we can rearrange this equation to make 𝑑𝑦 by 𝑑𝑥 the
subject — that is we can write 𝑑𝑦 by 𝑑𝑥 in terms of 𝑥 and 𝑦. The first thing to do here is to
factor out 𝑑𝑦 by 𝑑𝑥 in the two terms where it appears. Then, we can subtract eight 𝑥
minus five from both sides and then we just have something times 𝑑𝑦 by 𝑑𝑥 on the
left. And dividing through by this
something, we get that 𝑑𝑦 by 𝑑𝑥 is negative eight 𝑥 plus five over 16𝑦 minus
This is our final answer. Note that it is written in terms of
both 𝑥 and 𝑦. You might feel slightly uneasy
about this if you’re accustomed to writing 𝑑𝑦 by 𝑑𝑥 in terms of 𝑥 alone. But to write 𝑑𝑦 by 𝑑𝑥 in terms
of 𝑥 alone, we’d have to write 𝑦 in terms of 𝑥. And as we discussed at the start of
the video, that’s a bit tricky. And not only that, we’d probably
find that the expression for 𝑦 in terms of 𝑥 is quite ugly. And so we get quite an ugly
expression for 𝑑𝑦 by 𝑑𝑥.
So for this question and for
questions involving implicit differentiation more generally — that is when you
differentiate both sides with respect to a variable — it’s best to leave the answer
in terms of both 𝑥 and 𝑦.