_{Up Learn – A Level maths (edexcel) – Differentiating Combined Functions}

_{Up Learn – A Level maths (edexcel) – Differentiating Combined Functions}

**The Quotient Rule**

**It’s possible to differentiate a quotient of two functions without simplifying first. To do it, we use the quotient rule.**

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### More videos on Differentiating Combined Functions:

^{Introduction to Differentiating Combined Functions (free trial)}

^{The Chain Rule (free trial)}

^{The Power of the Chain Rule (free trial)}

^{Making Functions Composite Before Differentiating (free trial)}

^{The Chain Rule in Leibniz’s Notation (free trial)}

^{Finding Your Inner Function (free trial)}

^{Algebraic Inner Functions (free trial)}

^{Trigonometric Inner Functions (free trial)}

^{Exponential Inner Functions (free trial)}

^{Logarithmic Inner Functions (free trial)}

^{The Product Rule (free trial)}

^{The Quotient Rule}

## Calculus II

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2. What is a limit?

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3. What is Infinity?

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4. Limits, and Infinity

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5. Are Limits Real?

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6. Limits, and the Cartesian Plane

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7. Trickster limits

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8. Limits From the Right, and the Left

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9. The Difference Between Horizontal and Vertical Asymptotes

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10. Limit Notation

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2. Mathematically Describing a Changing World

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3. How Can We Figure Out the Gradient of a Curve

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4. The Gradient Function

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5. The Symbol for the Gradient Function

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6. Finding the Gradient Function – Part 1

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7. Finding the Gradient Function – Part 2

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8. Finding the Gradient Function – Part 3

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9. Finding the Gradient Function – Part 4

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10. Finding the Gradient Function – Part 5

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11. Finding the Gradient Function – Part 6

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12. Matching a Curve and its Gradient Function

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13. Increasing, or Decreasing – Part 1

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14. Increasing, or Decreasing – Part 2

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15. Increasing, or Decreasing – Part 3

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16. Proving Differentiation Finds the Gradient Function – Part 1

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17. Proving Differentiation Finds the Gradient Function – Part 2

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18. Proving Differentiation Finds the Gradient Function – Part 3

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19. Proving Differentiation Finds the Gradient Function – Part 4

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20. Proving Differentiation Finds the Gradient Function – Part 5

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21. dy/dx Notation – Part 1

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22. dy/dx Notation – Part 2

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23. dy/dx Notation – Part 3

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24. dy/dx Notation – Part 4

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25. Equation of a Tangent to a Curve

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26. Equation of a Normal to a Curve

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27. Locating Stationary Points

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28. Identifying Stationary Points

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29. Sketching the Gradient Function – Part 1

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30. Sketching the Gradient Function – Part 2

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31. Sketching the Gradient Function – Part 3

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32. Sketching the Gradient Function – Part 4

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33. Higher-Order Derivatives

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34. Rate of change

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35. Identifying Stationary Points Using Second-Order Derivatives

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2. When the Old Method Breaks Down

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3. Differentiating ln(x)

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4. Differentiating Other Log Functions

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5. The Change of Base Formula

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6. Differentiating sin(kx)

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7. Differentiating cos(kx)

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8. Differentiating tan(kx)

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9. Differentiating cot(kx)

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10. Differentiating sec(kx)

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11. Differentiating cosec(kx)

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12. Solving Problems with Non-Algebraic Curves

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13. Rheya and the Trigadorian Cloud

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14. Prove it All Night

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15. Holes in Graphs

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16. Using Limits to Mend Holes

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17. Mending Holes: A Special Case

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18. Mending Holes: Another Special Case

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19. Proving the Gradient Function of sin(x) Part 1

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20. Proving the Gradient Function of sin(x) Part 2

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21. Proving the Gradient Function of cos(x) Part 1

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22. Proving the Gradient Function of cos(x) Part 2

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2. The Chain Rule

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3. The Power of the Chain Rule

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4. Making Functions Composite Before Differentiating

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5. The Chain Rule in Leibniz’s Notation

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6. Finding Your Inner Function

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7. Algebraic Inner Functions

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8. Trigonometric Inner Functions

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9. Exponential Inner Functions

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10. Logarithmic Inner Functions

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11. The Product Rule

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12. The Power of the Product Rule

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13. Splitting Functions into Products Before Differentiating

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14. The Product Rule in Leibniz’s Notation

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15. Finding Your Factors

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16. Using the Chain Rule and Product Rule Together

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17. The Quotient Rule

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18. Splitting Fractions Apart Before Differentiating

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19. The Quotient Rule in Leibniz’s Notation

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20. Just Use the Numerator and Denominator

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21. Using the Chain Rule and Quotient Rule Together

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22. Using the Product Rule and Quotient Rule Together

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23. Using All Three Rules Together

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24. Using All Three Rules Together

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25. Proving the Derivatives of Cosec and Sec

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26. Proving the Derivatives of Tan and Cot

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2. Explicit and Implicit Equations

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3. Rearranging Implicit Equations

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4. Differentiating Implicit Equations by Rearranging

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5. Differential Equations

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6. Rearranging Differential Equations

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7. Implicit Differentiation Part 1

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8. Implicit Differentiation Part 2

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9. The Chain Rule Allows us to Differentiate Implicitly

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10. Implicit Differentiation and the Product Rule

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11. Implicit Differentiation and Fractions

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12. Solving Problems with Implicit Differentiation

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13. Getting Two Different Gradient Functions

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2. Differentiating Parametric Equations

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3. Finding the Gradient Function in Parametric Form

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4. Finding Stationary Points Using a Parameter

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5. Finding Tangents and Normals Using a Parameter

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2. Convex and Concave Curves

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3. The Gradient Value at Convex and Concave Intervals

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4. Are Straight Lines Convex or Concave?

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5. Using the Second Derivative

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6. Intervals in Which Functions are Convex or Concave

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7. Revisiting Points of Inflection

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8. The Second Derivative at Points of Inflection

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9. Using the Second Derivative to find Points of Inflection

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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.

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