The Quotient Rule

We’ve now seen how to use the chain rule and the product rule.

For example, find the derivative of  this [y=8(3x-1)34] equation… [just chain rule example]

To differentiate this [y=8(3x-1)34] equation, we need to use the chain rule…

First, make this [3x-1] our inner function, giving us these two equations [y=8u34 and u=3x-1]

And then, differentiate…multiply, [click one time][when you differentiate you get 3, then multiply the 8 => du/dx = 3, then dy/dx = 3 8 u1/4  ] and substitute this [3x-1replace u with that] expression into this term [18(3x-1)-14]

Now, we’ve seen before that it’s possible to divide one function by another, to find the…

It’s possible to divide one function by another, to find what we call the quotient of those two functions.

For example, we can divide this [fx=5x] function by this [gx=3×2+5x-2] function [fx=5x and gx=3×2+5x-2] to get…

We can divide this [fx=5x] function by this  [gx=3×2+5x-2] function get this quotient [fxgx=5x3x2+5x-2]

Now, technically, we could already differentiate the quotient of these functions…

But we’d first have to decompose this [fxgx=5x3x2+5x-2] into partial fractions first… [57(3x-1)+107(x+2)]

And then use the chain rule to differentiate each of these, one at a time!

Fortunately, there’s a much better way of differentiating the quotient of two functions…

Which doesn’t require us to find an expression for the quotient first, and then differentiate that.

First, like with the chain and product rules, differentiate each of these two functions…

These are the derivatives of the functions [two originals at top, two derivatives beneath]. 

f(x)=5x g(x)=3×2+5x-2

f'(x)=5  g'(x)=6x+5

Second, multiply the denominator by this [first derivative]…

Multiplying this [g(x)=3×2+5x-2] by this [f'(x)=5] gives us this [15×2+25x-10] expression.

Next, multiply the numerator [f(x)=5x] by this [g'(x)=6x+5] [second derivative]…

Multiplying this [f(x)=5x] by this [g'(x)=6x+5] gives us this [30×2+25x] expression.

Now, take this [15×2+25x-10]  expression, the one with the original denominator [v], and subtract this [30×2+25x] expression…[click one time]][(15×2+25x-10)-(30×2+25x)]

And finally, divide by the original denominator…squared….  [(3×2+5x-2)2  so ( (15×2+25x-10)-(30×2+25x) ) / (3×2+5x-2)2 ]

This process is known as the ‘quotient rule’

Now, the quotient rule is a little trickier than the chain rule and the product rule, as there’s a bit more to remember…

First, since we’re subtracting here, the order of these matters…and we must start with the expression containing the original denominator [v]… [so we would have u’v -uv’ ]

And then, we also have to remember to divide by the original denominator…squared! [v^2 ]

But nonetheless, just like the other two, this rule works for any two functions!

Like the quotient  [fxgx] of these two functions [fx=ex and gx=sin x , fxgx], for example…

Before, we wouldn’t have been able to find the derivative of this [exsin x] at all!

But using the quotient rule, we can first differentiate each of the functions… [fx=ex,gx=sin x ]

These are the derivatives of the functions [two originals at top, two derivatives beneath]. 

f(x)=ex g(x)=sin(x)

f'(x)=ex g'(x)=cos(x)

Second, multiply the denominator [g(x)=sin(x)] by this [f'(x)=ex] [first derivative]…

Multiplying this [g(x)=sin(x)] by this [f'(x)=ex] gives us this [exsin(x)] expression.

Next, multiply the numerator [f(x)=ex] by this [g'(x)=cos(x)] [second derivative]…

Multiplying this [f(x)=ex] by this [g'(x)=cos(x)] gives us this [excos(x)]  expression.

Now, take this [exsin(x)]  expression – with the original denominator – and subtract this [excos(x)]  expression…

And then divide by the original denominator [g(x)=sin(x)]…squared…. [sin2(x)]

Subtracting this [exsin(x)] from this [excos(x)], and then dividing by the denominator squared [sin2(x)]…gives us this… [exsin(x)-excos(x)sin2(x)] [click one time]

Next, here are two more functions [f(x)=ln(x) and g(x)=2x+1]

To find the derivative of this [f(x)=ln(x)] function divided by this [g(x)=2x+1] one, we first differentiate both functions… 

We first differentiate both functions, like this.  [f(x)=ln(x)   => f'(x)=1/x  g(x)=2x+1=> g'(x)=2]

And so now, by first subtracting this [ln(x)] multiplied by this [2], from this [2x+1] multiplied by this [1/x], and then dividing by this [(2x+1) ] squared, find this quotient’s derivative… 2x+1x-2ln(x)(2x+1)2

First, multiply this [ln(x)] by this [2]… [same order as before]   [2ln(x)] 

Second, multiply this [2x+1]  by this [1/x]…  [2x+1x]

Next, subtract this [2ln(x)]  from this [2x+1x]    [2x+1x- 2ln(x)] [click one time]

And finally, divide by the denominator squared! [(2x+1)2 ]  [1x(2x+1)-2ln(x)(2x+1)2] [click one time]

And so this time, given these two functions [f(x)=e4x and g(x)=cos(x)], find the derivative of this quotient [e4xcos(x)]…

First, differentiate both functions… [f(x)=e4x=> f'(x)=4e4x  ,  g(x)=cos(x)  => g'(x)=-sin(x) ]

Then, multiply this [ e4x ] by this [-sin(x) ]… [same order as before]

And this [cos(x) ] by this [4e4x] …

Next, subtract this [-e4xsin(x) ] from this [4e4xcos(x) ] …[click two times]

Before finally dividing by this [cos(x)] squared!  [cos2(x)] [4e4xcos(x)+e4xsin(x)cos2(x)][click one time]

Finally, it is possible to prove the quotient rule from first principles, using the general form of the gradient function.

But, in your A Level exam, you won’t be asked to – so we’re not going to get into that here.

To sum up, it’s possible to differentiate a quotient of two functions – without first finding an expression for that quotient, and then differentiating.

For example, given any two functions, fx and g(x), we can find the derivative of this quotient [fxgx] by…  

We can find the derivative of their quotientproduct by first differentiating both functions…

Then multiplying this [f(x)] by this [g'(x)], and this [g(x)] by this [f'(x)]…

Before subtracting this [f(x)g'(x)] from this [g(x)f'(x)] [click one time], and finally dividing by the denominator squared. 

We call this…

We call this the quotient rule.