### Video Transcript

The Quotient Rule

In this video, we will learn how to

find the derivative of a function using the quotient rule. We will be looking at various

examples of how it can be used.

Consider the function π¦ is equal

to negative three π₯ squared minus two π₯ plus 17 over the square root of π₯.

If we wanted to find the derivative

of this function, there are various methods which we could take. We could divide the numerator by

the denominator and simply differentiate the resulting function. Alternatively, we could write the

fraction as a product and find the derivative using the product rule. There is also an alternative method

which we can use to find this derivative. And it requires no simplifying or

rewriting of the equation. We call it the quotient rule. The derivation of the quotient rule

is a bit long-winded for this video. So we wonβt be covering it

here.

The quotient rule says that given

two differentiable functions, π’ of π₯ and π£ of π₯, the derivative of their

quotient is given by π by dπ₯ of π’ of π₯ over π£ of π₯ is equal to π£ of π₯ times

π by dπ₯ of π’ of π₯ minus π’ of π₯ times π by dπ₯ of π£ of π₯ all over π£ of π₯

squared. We can write this a lot more

succinctly in prime notation. This gives us π’ over π£ prime is

equal to π£π’ prime minus π’π£ prime all over π£ squared.

I find an easy way to remember the

quotient rule is with a rhyme. The rhyme goes LO π HI minus HI π

LO over the square of whatβs below. Where HI is the numerator of the

rational function which weβre differentiating. And LO is the denominator of this

rational function. And the πs which are within the

rhyme show where we need to differentiate. So π HI will be the differential

of the numerator of our function. And π LO is the differential of

the denominator of our function. You may find it easier to remember

via another means. But feel free to use this method

too.

Weβre now ready to look at some

examples.

Find the first derivative of π¦ is

equal to 8π₯ plus five over three π₯ plus 22.

Here, we can see that our function

π¦ is a rational function. So we can find its derivative by

using the quotient rule. The quotient rule tells us that if

we differentiate the quotient of two functions, so π’ over π£, with respect to

π₯. Then itβs equal to π£ multiplied by

the differential of π’ with respect to π₯ minus π’ timesed by the differential of π£

with respect to π₯ all over π£ squared. In order to find the first

derivative of π¦, letβs start by labeling π’ and π£ from our equation. π’ will be equal to the numerator

of the function, so eight π₯ plus five. And π£ will be equal to the

denominator of the function, so thatβs three π₯ plus 22.

Next, we must find d by dπ₯ of π’

and d by dπ₯ of π£, or dπ’ by dπ₯ and dπ£ by dπ₯. π’ and π£ are both polynomials. So we can simply differentiate them

term by term. Writing π’ in terms of powers of

π₯, we can say that itβs equal to eight π₯ to the power of one plus five π₯ to the

power of zero. In order to differentiate, we

simply multiply by the power and decrease the power by one. For the first time, we multiply by

the power, so thatβs one, and decrease the power by one, to zero. Leaving us with one timesed by

eight π₯ to the power of zero. For the second term, we multiply by

the power, so thatβs zero, and decrease the power by one, to negative one. Giving us zero multiplied by five

π₯ to the negative one.

In the first term, π₯ to the power

of zero is just one. So this becomes eight. In the second term, weβre

multiplying by zero. So that term becomes zero. Therefore, we find that dπ’ by dπ₯

is equal to eight. We can use a similar method to find

dπ£ by dπ₯. And we find that itβs equal to

three. Now that we found dπ’ by dπ₯ and

dπ£ by dπ₯, weβre ready to use the quotient rule. We find that dπ¦ by dπ₯ is equal to

π£, which is three π₯ plus 22, multiplied by dπ’ by dπ₯, so thatβs eight, minus π’,

so thatβs eight π₯ plus five, multiplied by dπ£ by dπ₯, so thatβs three. And this is all over π£ squared, so

thatβs three π₯ plus 22 all squared.

Next, we can expand the

brackets. And then simplify to find that our

solution is that the first derivative of π¦ is equal to 161 over three π₯ plus 22

all squared.

Now, we will look at a slightly

more complex example.

Find the first derivative of the

function π¦ is equal to four π₯ squared plus five π₯ plus five all over four π₯

squared minus two π₯ plus three.

We can see that our function is a

rational function. Therefore, we can use the quotient

rule in order to find the derivative. The quotient rule tells us that π’

over π£ dash is equal to π£π’ dash minus π’π£ dash all over π£ squared. Where π’ is the numerator of our

function and π£ is the denominator. In our case, π’ is equal to four π₯

squared plus five π₯ plus five. And π£ is equal to four π₯ squared

minus two π₯ plus three. Now, we must find π’ prime and π£

prime. We do this by differentiating π’

and π£ with respect to π₯.

Since π’ and π£ are both polynomial

functions, we can find their derivatives by taking each term and multiplying the

term by the power of π₯. And then, decreasing the power of

π₯ by one. And doing this, we find that π’

prime is equal to eight π₯ plus five. And π£ prime is equal to eight π₯

minus two. Substituting these into our

formula, we find that the first derivative of π¦ or π¦ prime is equal to π£

multiplied by π’ prime minus π’ multiplied by π£ prime all over π£ squared. Now, the result here looks quite

daunting. However, we can still expand the

brackets and then simplify. This is what we obtain after

expanding the brackets in the numerator.

Our final step is to simplify the

numerator. Now, we have reached our

solution. Which is that the first derivative

of π¦ or π¦ prime is equal to negative 28π₯ squared minus 16π₯ plus 25 all over four

π₯ squared minus two π₯ plus three all squared.

Now, letβs look at a slightly

different type of question.

Suppose that π of π₯ is equal to

π₯ squared plus ππ₯ plus π all over π₯ squared minus seven π₯ plus four. Given that π of zero is equal to

one and π prime of zero is equal to four, find π and π.

Our first step in this question can

be to substitute π₯ equals zero into π of π₯. Since weβre given that π of zero

is equal to one. We obtain that π of zero is equal

to zero squared plus π times zero plus π all over zero squared minus seven times

zero plus four. Now, all of these terms will go to

zero apart from π and four. Weβre left with π of zero is equal

to π over four. Next, we use the fact that the

question has told us that π of zero is equal to one. And so, we can set this equal to

one. From this, we find that π is equal

to four. Next, we can use the fact that π

prime of zero is equal to four. However, first of all, we must find

π prime of π₯. In order to do this, we need to

differentiate π. Since π is a rational function, we

can use the quotient rule in order to find its derivative.

The quotient rule tells us that π’

over π£ prime is equal to π£ times π’ prime minus π’ times π£ prime all over π£

squared. Setting our function π of π₯ equal

to π’ over π£, we obtain that π’ is equal to π₯ squared plus ππ₯ plus π. And π£ is equal to π₯ squared minus

seven π₯ plus four. We can find π’ prime and π£ prime

by differentiating these two functions. Giving us that π’ prime is equal to

two π₯ plus π and π£ prime is equal to two π₯ minus seven. Now, we can substitute these into

the quotient rule. We obtain that π prime of π₯ is

equal to π₯ squared minus seven π₯ plus four multiplied by two π₯ plus π minus π₯

squared plus ππ₯ plus π multiplied by two π₯ minus seven all over π₯ squared minus

seven π₯ plus four all squared.

Now, we could simplify π prime of

π₯ at this point. However, weβre going to be

substituting in π₯ is equal to zero. And so, a lot of these terms would

just disappear. Letβs simply substitute π₯ equals

zero in here. We obtain this. However, a lot of the terms will

vanish to zero, which leaves us with four π plus seven π all over 16. Now, we have found that π is equal

to four earlier. And so, we can substitute this in,

giving us four π plus 28 all over 16.

Since the question has told us that

π prime of zero is equal to four, we can set this equal to four. Then, we simply rearrange this in

order to solve for π. Now, we obtain our solution that π

is equal to nine. Weβve now found the values of both

π and π, which completes the solution to this question.

In the next example, weβll be

looking at a slightly different type of question.

Let π of π₯ be equal to π of π₯

over negative four β of π₯ minus five. Given that π of negative two is

equal to negative one, π prime of negative two is equal to negative eight, β of

negative two is equal to negative two, and β prime of negative two is equal to five,

find π prime of negative two.

In this question, weβre asked to

find π prime of negative two. So letβs start by differentiating

π of π₯. π of π₯ is a rational function, so

weβll need to use the quotient rule. The quotient rule tells us that π’

over π£ prime is equal to π£π’ prime minus π’π£ prime all over π£ squared. Setting π of π₯ equal to π’ over

π£, we can see that π’ is equal to π of π₯. And π£ is equal to negative four β

of π₯ minus five. π’ prime will be equal to π of π₯

prime. Now, the prime simply represents a

differentiation with respect to π₯. So therefore, π of π₯ prime is

identical to π prime of π₯. Next, we need to find π£ prime. So thatβs negative four β of π₯

minus five prime.

Now, again, since a prime simply

represents a differentiation with respect to π₯, we can apply normal differentiation

rules here. And so, differentiating the

constant term negative five will result in zero. So we can say that this is equal to

negative four β of π₯ prime. Now, since our function β of π₯ is

being multiplied by a constant, negative four. We can use our derivative rules and

take the negative four out of the derivative. Giving us negative four multiplied

by β of π₯ prime.

And now, we can apply the same

logic as we did for π of π₯ prime. And we can say that π£ prime is

equal to negative four β prime of π₯. Now, we can substitute into the

quotient rule in order to find π prime of π₯. Now that we have found π prime of

π₯, we can substitute in π₯ is equal to negative two. Now, we have formed an equation in

terms of π of negative two, π prime of negative two, β of negative two, and β

prime of negative two. All of which we have been given the

value of in the question. And so, weβre able to substitute in

these values here.

Now, our final step in finding π

prime of negative two is to simplify this. Expanding the brackets, we get

negative 24 minus 20 all over nine. This gives us a solution that π

prime of negative two is equal to negative 44 over nine.

Next, we will see how we can

differentiate a function which consists of two rational expressions.

If π¦ is equal to π₯ plus five over

π₯ minus five minus π₯ minus five over π₯ plus five, find dπ¦ by dπ₯.

Our function, π¦, consists of two

rational expressions, π₯ plus five over π₯ minus five and π₯ minus five over π₯ plus

five. And we could find dπ¦ by dπ₯ by

using the quotient rule on these two rational expressions. However, this would require using

the quotient rule twice. We can make our work a little

easier by combining the two rational expressions into one. We obtain that π¦ is equal to π₯

plus five squared minus π₯ minus five squared all over π₯ minus five times π₯ plus

five. We can expand the brackets and then

simplify to obtain that π¦ is equal to 20π₯ over π₯ squared minus 25.

And now, our function consists of

only one rational expression. Weβre ready to use the quotient

rule to differentiate this function. The quotient rule tells us that π’

over π£ prime is equal to π£π’ prime minus π’π£ prime over π£ squared. Setting π¦ equal to π’ over π£, we

obtain that π’ is equal to 20π₯ and π£ is equal to π₯ squared minus 25. Next, we can find π’ prime and π£

prime, which gives us that π’ prime is equal to 20 and π£ prime is equal to two

π₯.

Now, we can substitute them into

the quotient rule in order to find that dπ¦ by dπ₯ is equal to π₯ squared minus 25

times 20 minus 20π₯ times two π₯ all over π₯ squared minus 25 squared. We simplify this to obtain that dπ¦

by dπ₯ is equal to negative 20π₯ minus 500 all over π₯ squared minus 25 squared.

In our final example, we will see

how to evaluate the derivative of a rational function at a point.

Evaluate π prime of three, where

π of π₯ is equal to π₯ over π₯ plus two minus π₯ minus three over π₯ minus two.

Now, our function is the difference

of two rational expressions. We can start by combining the two

rational expressions into one. We obtain that π of π₯ is equal to

negative π₯ plus six all over π₯ squared minus four. And we have written π as a

rational function. And weβre ready to use the quotient

rule, which tells us that π’ over π£ prime is equal to π£π’ prime minus π’π£ prime

all over π£ squared. Setting π of π₯ equal to π’ over

π£, we obtain that π’ is equal to negative π₯ plus six, and π£ is equal to π₯

squared minus four. We then find that π’ prime is equal

to negative one, and π£ prime is equal to two π₯.

Substituting π’, π£, π’ prime, and

π£ prime back into the quotient rule. We find that π prime of π₯ is

equal to π₯ squared minus four multiplied by negative one minus negative π₯ plus six

multiplied by two π₯ all over π₯ squared minus four squared. In order to find π prime of three,

we simply substitute π₯ equals three into π prime of π₯. We obtain that π dash of three is

equal to three squared minus four multiplied by negative one minus negative three

plus six multiplied by two times three all over three squared minus four

squared. Which simplifies to negative five

minus 18 over 25. This gives us a solution that π

prime of three is equal to negative 23 over 25.

Now, we have seen a variety of

examples of the quotient rule. Letβs cover some key points of the

video. To find the derivative of the

quotient of two differentiable functions, π’ of π₯ and π£ of π₯, we can use the

quotient rule which states that d by dπ₯ of π’ of π₯ over π£ of π₯ is equal to π£ of

π₯ times d by dπ₯ of π’ of π₯ minus π’ of π₯ times d by dπ₯ of π£ of π₯ all over π£

of π₯ squared. This is often written more

succinctly using prime notation, as follows. π’ over π£ prime is equal to π£π’

prime minus π’π£ prime all over π£ squared. Before applying the quotient rule,

it is worth checking whether itβs possible to simplify the expression for the

function. This is particularly relevant when

the function is expressed as the sum or difference of two rational expressions.