### Video Transcript

In this video, weβll learn how to use
implicit differentiation to help us find the derivative of functions expressed
implicitly as functions of π₯. The majority of differentiation
problems you will have come across so far will have involved functions written
explicitly as functions of π₯, such as π¦ equals three π₯ squared sin π₯. In this video, weβre going to
explore how implicit differentiation, which is an extension of the chain rule,
allows us to easily differentiate equations such as the Cartesian equation of a
circle, π₯ squared plus π¦ squared equals one for example. Weβll also explore what this means
for the second derivative and higher order derivatives.

The chain rule allows us to
differentiate composite functions. It says that for two differentiable
functions, π and β, such that π is the composite function π of β of π₯, the
derivative of π is the derivative of β of π₯ times the derivative of π evaluated
at β of π₯. Itβs often more intuitive though to
write this as dπ¦ by dπ₯ equals dπ’ by dπ₯ times dπ¦ by dπ’, where π¦ is a function
of π’ and π’ is a function of π₯. Letβs demonstrate how this chain
rule can help us to find the derivative of an implicit function.

Consider the equation π₯ squared
plus π¦ squared equals one. Using implicit differentiation,
find an expression for dπ¦ by dπ₯ in terms of π₯ and π¦. And thereβs a second part to this
question. For the semicircle where π¦ is
greater than or equal to zero, express π¦ explicitly in terms of π₯, then
differentiate this expression to get an expression for dπ¦ by dπ₯ in terms of
π₯.

Weβll begin with part one. To differentiate the function
implicitly, weβre going to begin by differentiating both sides of the equation with
respect to π₯. This is probably going to look a
little strange, but bear with me. We say that the derivative of π₯
squared plus π¦ squared with respect to π₯ is equal to the derivative of one with
respect to π₯. And we absolutely must do this to
both sides of the equation. Next, we differentiate what we
can. Itβs quite straightforward to
differentiate one with respect to π₯. Itβs simply zero. But how do we differentiate π₯
squared plus π¦ squared?

Well, the derivative of π₯ squared
is two π₯. But the derivative of π¦ squared is
a little more strange. We know that π¦ squared is a
function of π¦. And, in turn, π¦ is a function of
π₯. So here, we can use the chain
rule. We say that the derivative of π¦
squared with respect to π₯ is equal to the derivative of π¦ squared with respect to
π¦ times the derivative of π¦ with respect to π₯. Well, the derivative of π¦ squared
with respect to π¦ is two π¦. And the derivative of π¦ with
respect to π₯ is simply dπ¦ by dπ₯. So our equation now becomes two π₯
plus two π¦ dπ¦ by dπ₯ equals zero.

Remember, we want an equation for
the derivative. So weβre going to make dπ¦ by dπ₯
the subject by first subtracting two π₯ from both sides of our equation. That gives us two π¦ dπ¦ by dπ₯ is
equal to negative two π₯. Next, weβre going to divide through
by two π¦. And we see that dπ¦ by dπ₯ is equal
to negative two π₯ divided by two π¦. Well, the twos cancel. And we found an expression for dπ¦
by dπ₯ in terms of π₯ and π¦; itβs negative π₯ over π¦.

For part two of this question,
weβre going to go back to our original equation. And weβre going to begin by making
π¦ the subject of the equation. Thatβs expressing π¦ explicitly in
terms of π₯. We can subtract π₯ squared from
both sides. And then we take the square root of
both sides of the equation. Now, usually, we take by the
positive and negative square root. Here though, weβre told the
semicircle is such that π¦ is greater than or equal to zero. So weβre just going to consider the
positive square root. And we have an explicit function in
π₯.

Now, writing it as one minus π₯
squared to the power of a half, we can see that we can use the general power rule to
differentiate this. This says that if π’ is some
function in π₯, the derivative of π’ to the power of π with respect to π₯ is equal
to π times π’ to the power of π minus one multiplied by the derivative of π’ with
respect to π₯. And this is of course when π is a
real number. So the derivative of one minus π₯
squared to the power of a half is a half times one minus π₯ squared to the power of
negative half multiplied by dπ’ by dπ₯. But, actually, π’ is equal to one
minus π₯ squared. So the derivative of π’ with
respect to π₯ is just negative two π₯. Once again, we have some twos to
cancel. And our expression for dπ¦ by dπ₯
in terms of π₯ is negative π₯ over the square root of one minus π₯ squared.

Notice that since we said that π¦
was equal to the square root of one minus π₯ squared, we could have written this as
dπ¦ by dπ₯ equals negative π₯ over π¦. And thatβs the same answer as we got
in part one of this question. And, of course, we must remember that
π¦ cannot be equal to zero here. Now, this example demonstrates some
important points. Firstly, even though it was easy
enough to express this relation as an explicit function, it was simpler to
differentiate using implicit differentiation.

And, in general, when weβre
differentiating implicitly, we can use the following version of the chain rule. This says the derivative of a
function in π¦ with respect to π₯ is equal to the derivative of that function in π¦
with respect to π¦ times dπ¦ by dπ₯. Itβs really useful to commit this
version of the chain rule to memory. Letβs now apply it to a more
complicated example.

Find dπ¦ by dπ₯ by implicit
differentiation if negative π of to the π¦ sin π₯ is equal to four π₯π¦ plus two
π₯.

To differentiate this function
implicitly, we begin by differentiating both sides of the equation with respect to
π₯. We begin by writing this as d by dπ₯
of negative π to the π¦ times sin π₯ equals d by dπ₯ of four π₯π¦ plus two π₯. Letβs differentiate each bit with
respect to π₯. Well, the derivative of two π₯ is
straightforward; itβs simply two. Weβre going to have to use the
product rule combined with the chain rule to differentiate four π₯π¦ and negative π
to the π¦ sin π₯. The special version of the chain
rule that we need says the derivative of π of π¦ with respect to π₯ is equal to the
derivative of π¦ with respect to π¦ times dπ¦ by dπ₯. And the product rule says that the
derivative of π’ times π£ is equal to π’ times the derivative of π£ plus π£ times
the derivative of π’.

Weβll begin by differentiating four
π₯π¦. Weβre going to let π’ be equal to
four π₯ and π£ be equal to π¦. Then the derivative of π’ with
respect to π₯ is simply four. The derivative of π£ with respect
to π₯ is equal to the derivative of π¦ with respect to π¦ which is one times dπ¦ by
dπ₯ which is simply dπ¦ by dπ₯. π’ times dπ£ by dπ₯ is then four π₯
multiplied by dπ¦ by dπ₯. And π£ multiplied by dπ’ by dπ₯ is
equal to π¦ times four. And, therefore, the right-hand side
of our equation is four π₯ dπ¦ by dπ₯ plus four π¦ plus two. Letβs now repeat this process for
negative π to the power of π¦ times sin π₯.

This time, letβs say π’ is equal to
negative π to the power of π¦ and π£ is equal to sin π₯. The derivative of sin π₯ is cos
π₯. And then the derivative of π’ with
respect to π₯ is equal to the derivative of negative π to the π¦ with respect to
π¦, thatβs negative π to the π¦, times dπ¦ by dπ₯. And then when we use the product
rule, we see that dπ¦ by dπ₯ is equal to negative π to the π¦ cos π₯ minus π to
the π¦ sin π₯ dπ¦ by dπ₯. So our equation is now as
shown. Remember, weβre trying to find an
equation for dπ¦ by dπ₯. So weβre going to rearrange and
make dπ¦ by dπ₯ the subject. When we do, we see that negative π
to the π¦ cos π₯ minus four π¦ minus two is equal to four π₯ plus π to the π¦ sin
π₯ all multiplied by dπ¦ by dπ₯.

Now, we can actually factor negative
one on the left-hand side. And then, we divide through by four
π₯ plus π to the π¦ sin π₯. And we see that dπ¦ by dπ₯ is equal
to negative π to the π¦ cos π₯ plus four π¦ plus two over four π₯ plus π to the π¦
sin π₯. And, of course, this derivative
stands when the denominator is not equal to zero, when four π₯ plus π to the π¦ sin
π₯ is not equal to zero. Itβs fairly common to use implicit
differentiation to find the equation of a tangent to a curve defined implicitly. In our next example, weβll look at
how we can use implicit differentiation to address a problem of this type.

The equation π¦ squared minus 24π₯
cubed plus 24π₯ equals zero describes a curve in the plane. 1) Find the coordinates of two
points on this curve, where π₯ equals negative one-half. 2) Determine the equation of the
tangent at the points where π₯ equals negative one-half and the π¦-coordinate is
positive. 3) Find the coordinates of another
point, if it exists, at which the tangent meets the curve.

For part one, to find the points
where π₯ is equal to negative one-half, weβre going to substitute this value of π₯
into our equation and solve for π¦. Thatβs π¦ squared minus 24 times
negative a half cubed plus 24 times negative a half. And thatβs equal to zero. When we evaluate this, we get π¦
squared plus three minus 12 equals zero or π¦ squared minus nine equals zero. Weβre going to solve this equation
by adding nine to both sides.

And our last step is to find the
square root of both sides of this equation, remembering to take by the positive and
negative square roots of nine. The square root of nine is
three. So we see that π¦ is equal to three
and negative three, when π₯ is equal to negative one-half. In coordinate form, thatβs negative
a half, three and negative a half, negative three.

We now consider part two. We need to find the equation of the
tangent at a point where π₯ equals negative a half and the π¦-coordinate is
positive. Thatβs the coordinate negative a
half, three. Weβll first, however, need to find
the gradient of the tangent to the curve. This will be the derivative of the
equation for the curve evaluated at π₯ equals negative a half and π¦ equals
three. So weβre going to differentiate our
equation implicitly. Thatβs d by dπ₯ of π¦ squared minus
24π₯ cubed plus 24π₯ equals d by dπ₯ of zero. The derivative of π¦ squared is the
derivative of π¦ squared with respect to π¦ times the derivative of π¦ with respect
to π₯. Itβs two π¦ dπ¦ by dπ₯. The derivative of negative 24π₯
cubed is three times negative 24π₯ squared. Thatβs negative 72π₯ squared. The derivative of 24π₯ is 24. And the derivative of zero is
zero.

So we see that two π¦ dπ¦ by dπ₯
minus 72π₯ squared plus 24 is equal to zero. We need an equation for dπ¦ by
dπ₯. So weβre going to rearrange to make
dπ¦ by dπ₯ the subject. We add 72π₯ squared to both sides of
this equation and subtract 24. Next, weβre going to divide through
by two π¦. And we see that dπ¦ by dπ₯ is equal
to 72π₯ squared minus 24 over two π¦, which simplifies to 36π₯ squared minus 12 over
π¦. Remember, we want to find the
gradient of the tangent of the curve at negative a half, three. So weβre going to substitute π₯
equals negative a half and π¦ equals three into the equation for the derivative. When we do, we see that the
gradient of our tangent is 36 times negative a half squared minus 12 all over three,
which is negative one.

Finally, we substitute what we know
about our tangent into the equation of a straight line. And we get π¦ minus three equals
negative one times π₯ minus negative a half. Distributing the parentheses and
simplifying then rearranging for π¦, we get π¦ equals five over two minus π₯.

And now we consider part three. We need to find a point, if it
exists, where this tangent meets the curve again. We therefore want to solve
simultaneously the equations π¦ squared minus 24π₯ cubed plus 24π₯ equals zero and
π¦ equals five over two minus π₯. We can achieve this by substituting
π¦ equals five over two minus π₯ into the original equation for the curve. That gives us five over two minus
π₯ all squared minus 24π₯ cubed plus 24π₯ equals zero. Expanding the parentheses and
multiplying through by negative one, and we end up with the equation shown.

Next, we could solve this using a
scientific calculator. Alternatively, we know that π₯
equals negative a half is a root to this equation. And we can therefore factor out π₯
plus a half. And we can do this by using long
division or matching coefficients. The first step towards equating
coefficients would be to write the equation shown. And whilst itβs outside the scope
of this video to spend a lot of time performing this, we should find that π is
equal to 24, π is negative 13, and π is negative 25 over two. And you can pause the video here if
you like and see if you can complete that step yourself.

Our last step is to solve the
quadratic equation 24π₯ squared minus 13π₯ minus 25 over two. And we could do that by using the
quadratic formula or completing the square. When we solve this, we see that π₯
is equal to negative a half. So thatβs a repeated root. And π₯ is equal to 25 over 24. Weβre finding the coordinates, so
we substitute π₯ equals 25 over 24 into the equation five over two minus π₯. And that gives us a π¦-value of 35
over 24. And so the tangent with equation π¦
equals five over two minus π₯ meets the curve at negative a half, three and
twenty-five twenty-fourths, thirty-five twenty-fourths. In our very final example, weβll
look at how implicit differentiation can help us to find higher order
derivatives.

Given that two sin π¦ minus five
cos π₯ equals negative four, determine the second derivative of π¦ by implicit
differentiation.

To find to the second derivative of
π¦, sometimes called π¦ double prime, weβre going to need to differentiate our
function twice. Notice how the function itself is
expressed implicitly as functions of π₯. So weβre going to use implicit
differentiation and begin by finding the derivative of both sides. Now the derivative of negative four
with respect to π₯ is fairly easy to work out. Itβs just zero. Similarly, we can find the
derivative of negative five cos π₯ with respect to π₯. Thatβs five sin π₯. But what about the derivative of
two sin π¦ with respect to π₯? Well, here we use a special version
of the chain rule. This says that the derivative of a
function in π¦ with respect to π₯ is equal to the derivative of that function with
respect to π¦ times dπ¦ by dπ₯.

Well, the derivative of two sin π¦
with respect to π¦ is two cos π¦. So our equation is two cos π¦ dπ¦ by
dπ₯ plus five sin π₯. And thatβs equal to zero. We can find the first derivative
then by subtracting five sin π₯ from both sides of the equation and dividing through
by two cos π¦. Remember, though, we were looking
to find the second derivative. So here weβre going to need to use
the quotient rule to differentiate negative five sin π₯ over two cos π¦. According to our notation, we can
let π’ be equal to negative five sin π₯ and π£ equal two cos π¦. dπ’ by dπ₯ is
negative five cos π₯. Then since the derivative of π£
with respect to π¦ is negative two sin π¦, we can see that dπ£ by dπ₯ is negative
two sin π¦ dπ¦ by dπ₯.

We substitute each of these into
the formula for the quotient rule. We then simplify and spot that we
said that dπ¦ by dπ₯ is negative five sin π₯ over two cos π¦. So we can substitute this into the
formula for the second derivative. To simplify this somewhat, we
multiply both the numerator and denominator of our fraction by two cos π¦. And weβre going to look to simplify
just a little more. Distributing this negative one, we
can see that the second derivative of our function is 25 sin squared π₯ sin π¦ minus
10 cos π₯ cos squared π¦ all over cos cubed π¦. And this formula is valid as long
as cos π¦ is not equal to zero.

In this video, weβve seen that when
given a function defined implicitly, we can use a special version of the chain rule
to differentiate it. The derivative of a function in π¦
with respect to π₯ is equal to the derivative of that function π¦ with respect to π¦
times dπ¦ by dπ₯. We also saw that if it was possible
to rewrite a relation as an explicit function, it was sometimes simpler to use
implicit differentiation. We also saw that when we
differentiate implicitly, weβll get an expression for dπ¦ by dπ₯ in terms of both π₯
and π¦. We also saw that we can find higher
order derivatives by using implicit differentiation. And in these cases, weβll need to
substitute expressions for lower order derivatives to help us simplify the
expressions.