A multiple-choice quiz
by jonnowales.
Estimated time: 5 mins.

- Home
- »
- Quizzes
- »
- Science Trivia
- »
- Math
- »
- Calculus

Scroll down to the bottom for the answer key.

Quiz Answer Key and Fun Facts

Answer:
**dy/dx = 2x**

If you differentiate an equation such as y = x^2 (^ means 'raised to the power of') once you are essentially measuring the change in 'y', with respect to 'x'. The change in 'y', with respect to 'x' is represented by dy/dx which is usually said as "d-y-d-x".

When differentiating a straightforward equation such as y = x^2, you simply lower the value of the exponent, or power, by one and multiply by the original value of the exponent. For example, the exponent in the equation y = x^2 is '2', decrease this by one and you are left with dy/dx = x^1 which can be expressed as just dy/dx = x. The original value of the exponent was 2, so multiply the base (in this case 'x') by two. The result is dy/dx = 2x.

If you differentiate an equation such as y = x^2 (^ means 'raised to the power of') once you are essentially measuring the change in 'y', with respect to 'x'. The change in 'y', with respect to 'x' is represented by dy/dx which is usually said as "d-y-d-x".

When differentiating a straightforward equation such as y = x^2, you simply lower the value of the exponent, or power, by one and multiply by the original value of the exponent. For example, the exponent in the equation y = x^2 is '2', decrease this by one and you are left with dy/dx = x^1 which can be expressed as just dy/dx = x. The original value of the exponent was 2, so multiply the base (in this case 'x') by two. The result is dy/dx = 2x.

Answer:
**Gradient (slope)**

As the line on the graph is a curve, the gradient will be different for different values of 'x'. If the graph showed a straight line then it would be possible to get an equation of the form 'y = mx + c' where 'm' is the constant gradient. With a curve it isn't quite as simple to obtain the gradient because you can't just take the value of 'm' from the equation as it doesn't exist! In order to obtain the gradient at a specific point on a curve you first have to differentiate the equation of the curve as is shown in the question. This new expression is known as the derivative and it is the derivative of 'y' in this case, with respect to 'x'. Then if you wish to find out the gradient at the point on the curve where 'x' equalled say 3, you just put 'x = 3' into the derivative.

Thus for the curve 'y = x^3 + 2x^2 + 4' you can obtain the derivative 'dy/dx = 3x^2 + 4x' and the gradient at the point 'x = 3' is:

dy/dx = 3(3)^2 + 4(3)

dy/dx = 27 + 12

dy/dx = gradient = 39.

If the gradient is 39 then you would expect to see a steep incline of the curve at that point.

As the line on the graph is a curve, the gradient will be different for different values of 'x'. If the graph showed a straight line then it would be possible to get an equation of the form 'y = mx + c' where 'm' is the constant gradient. With a curve it isn't quite as simple to obtain the gradient because you can't just take the value of 'm' from the equation as it doesn't exist! In order to obtain the gradient at a specific point on a curve you first have to differentiate the equation of the curve as is shown in the question. This new expression is known as the derivative and it is the derivative of 'y' in this case, with respect to 'x'. Then if you wish to find out the gradient at the point on the curve where 'x' equalled say 3, you just put 'x = 3' into the derivative.

Thus for the curve 'y = x^3 + 2x^2 + 4' you can obtain the derivative 'dy/dx = 3x^2 + 4x' and the gradient at the point 'x = 3' is:

dy/dx = 3(3)^2 + 4(3)

dy/dx = 27 + 12

dy/dx = gradient = 39.

If the gradient is 39 then you would expect to see a steep incline of the curve at that point.

Answer:
**f '(x)**

In fact f'(x) is widely used in modern mathematics to indicate that the expression you are looking at is a derivative of a function. Despite the widespread use of Leibniz's notation (dy/dx) for the differentiation of equations of the form 'y = ax^n + bx^n...', his notation can become a little less pleasing to look at when the derivative in question is that of a function such as f(x). The apostrophe used in the Lagrange notation for the derivative of a function, f'(x), is known as a prime. In speech it is generally said "f prime of x" but you may often hear "f dashed of x" or variations on that theme.

(A space was put in between f and the prime mark in the answer for clarity).

In fact f'(x) is widely used in modern mathematics to indicate that the expression you are looking at is a derivative of a function. Despite the widespread use of Leibniz's notation (dy/dx) for the differentiation of equations of the form 'y = ax^n + bx^n...', his notation can become a little less pleasing to look at when the derivative in question is that of a function such as f(x). The apostrophe used in the Lagrange notation for the derivative of a function, f'(x), is known as a prime. In speech it is generally said "f prime of x" but you may often hear "f dashed of x" or variations on that theme.

(A space was put in between f and the prime mark in the answer for clarity).

Answer:
**cos(x)**

When first learning the list of derivatives of trigonometric functions one could be forgiven for thinking that the subject is a little tedious but the mathematics behind the list is beautiful. If you find a source which shows you the proof behind the knowledge that the derivative of 'sin(x)' is 'cos(x)' then it is well worth having a look!

If one has a graph with a curve 'y = sin(x)' then a sine wave will be seen which is a curve that starts at the origin, rises to 'y = 1' at the point 'x = 90 degrees' (or in radians, pi/2) and then sinks to 'y = 0' again at 'x = 180 degrees' (or pi radians). This is repeated every 180 degrees but there is one difference. Once a 180 degree cycle has passed where the curve rises to 'y = 1' then the next 180 degree cycle will show the curve going down to 'y = -1'. This pattern alternates until your patience runs out and you can't be bothered to draw the graph anymore! Mathematically of course it can repeat until infinity.

When the sine curve is differentiated it almost magically transforms into a cosine curve which is a wonderful example of how maths works! If you'd like to see diagrammatic representations of the sine curve and the cosine curve then the internet will not disappoint.

When first learning the list of derivatives of trigonometric functions one could be forgiven for thinking that the subject is a little tedious but the mathematics behind the list is beautiful. If you find a source which shows you the proof behind the knowledge that the derivative of 'sin(x)' is 'cos(x)' then it is well worth having a look!

If one has a graph with a curve 'y = sin(x)' then a sine wave will be seen which is a curve that starts at the origin, rises to 'y = 1' at the point 'x = 90 degrees' (or in radians, pi/2) and then sinks to 'y = 0' again at 'x = 180 degrees' (or pi radians). This is repeated every 180 degrees but there is one difference. Once a 180 degree cycle has passed where the curve rises to 'y = 1' then the next 180 degree cycle will show the curve going down to 'y = -1'. This pattern alternates until your patience runs out and you can't be bothered to draw the graph anymore! Mathematically of course it can repeat until infinity.

When the sine curve is differentiated it almost magically transforms into a cosine curve which is a wonderful example of how maths works! If you'd like to see diagrammatic representations of the sine curve and the cosine curve then the internet will not disappoint.

Answer:
**Chain rule**

The chain rule is used when there is the need to differentiate a two function composite. In the example above 'y = sin(4x)', the two functions are 'sin(u)' and '4x' where 'u = 4x'. (The letter 'u' was chosen randomly to allow for the easy expression of the first function). All the chain rule simply states is that you differentiate the first function and multiply the newly obtained derivative by the derivative of the second function. The way one decides which function is to be differentiated first is a bit like selecting which of the sets of cutlery to use at a very formal dinner. You start with the outermost function and work your way in.

So, how does the chain rule apply to 'y = sin(4x)'? Well, firstly differentiate the sine part of the equation and that is simply cosine. So after the first of the two differentiation processes we have 'dy/dx = cos(4x)', but, the second function, '4x', also needs to be differentiated and then multiplied to the first derivative. If you differentiate '4x' you simply get four. Thus, the chain rule gives us the aesthetically pleasing result of 'dy/dx = 4.cos(4x)'.

The chain rule is used when there is the need to differentiate a two function composite. In the example above 'y = sin(4x)', the two functions are 'sin(u)' and '4x' where 'u = 4x'. (The letter 'u' was chosen randomly to allow for the easy expression of the first function). All the chain rule simply states is that you differentiate the first function and multiply the newly obtained derivative by the derivative of the second function. The way one decides which function is to be differentiated first is a bit like selecting which of the sets of cutlery to use at a very formal dinner. You start with the outermost function and work your way in.

So, how does the chain rule apply to 'y = sin(4x)'? Well, firstly differentiate the sine part of the equation and that is simply cosine. So after the first of the two differentiation processes we have 'dy/dx = cos(4x)', but, the second function, '4x', also needs to be differentiated and then multiplied to the first derivative. If you differentiate '4x' you simply get four. Thus, the chain rule gives us the aesthetically pleasing result of 'dy/dx = 4.cos(4x)'.

Answer:
**sec^2 (x)**

The proof behind this starts with the fact that 'tan(x) = sin(x)/cos(x)'. This new expression can be differentiated by the quotient rule which is 'dy/dx = [(v.du/dx) - (u.dv/dx)]/v^2' where 'v = cos(x)' and 'u = sin(x)'. This will give a result of 'dy/dx = cos(x)cos(x) + sin(x)sin(x))/cos^2 (x)'. However, a very useful trigonometric identity is 'cos(x)cos(x) + sin(x)sin(x) = 1'. So, the whole top line can be just replaced by one. Thus:

'dy/dx = 1/cos^2 (x)' and the expression to the right of the equal sign is 'sec^2 (x)'! This is short for secant squared of 'x' and this is the inverse trigonometric function of cosine squared of 'x'.

The proof behind this starts with the fact that 'tan(x) = sin(x)/cos(x)'. This new expression can be differentiated by the quotient rule which is 'dy/dx = [(v.du/dx) - (u.dv/dx)]/v^2' where 'v = cos(x)' and 'u = sin(x)'. This will give a result of 'dy/dx = cos(x)cos(x) + sin(x)sin(x))/cos^2 (x)'. However, a very useful trigonometric identity is 'cos(x)cos(x) + sin(x)sin(x) = 1'. So, the whole top line can be just replaced by one. Thus:

'dy/dx = 1/cos^2 (x)' and the expression to the right of the equal sign is 'sec^2 (x)'! This is short for secant squared of 'x' and this is the inverse trigonometric function of cosine squared of 'x'.

Answer:
**Product rule**

It is standard practise in mathematics to use a full stop (period) to indicate a multiplication when the letter 'x' is being used algebraically. This is to stop any confusion as to what the letter 'x' actually means at any given time. The product rule in word form states that one keeps the first term and multiplies it by the derivative of the second term.

This is then added to the result of the multiplication between the derivative of the first term and the original second term. If we take the example 'y = x.sin(x)', the first term is 'x' and the second term is 'sin(x)'. To differentiate this equation you keep the first term, 'x', the same and multiply it by the derivative of the second term, 'sin(x)', and 'sin(x)' differentiated is 'cos(x)'. Thus the multiplication of these two terms leads to a result of 'x.cos(x)'.

Then, simply keep the second term the same and multiply it to the derivative of the first term. The derivative of the first term, 'x', in this case is just one. Thus the multiplication of these two terms leads to a result of '1.sin(x)'.

The product rule states that the newly ascertained expressions should be added together and therefore the derivative of 'y = x.sin(x)' is 'dy/dx = x.cos(x) + sin(x)'. Cool stuff, eh?

It is standard practise in mathematics to use a full stop (period) to indicate a multiplication when the letter 'x' is being used algebraically. This is to stop any confusion as to what the letter 'x' actually means at any given time. The product rule in word form states that one keeps the first term and multiplies it by the derivative of the second term.

This is then added to the result of the multiplication between the derivative of the first term and the original second term. If we take the example 'y = x.sin(x)', the first term is 'x' and the second term is 'sin(x)'. To differentiate this equation you keep the first term, 'x', the same and multiply it by the derivative of the second term, 'sin(x)', and 'sin(x)' differentiated is 'cos(x)'. Thus the multiplication of these two terms leads to a result of 'x.cos(x)'.

Then, simply keep the second term the same and multiply it to the derivative of the first term. The derivative of the first term, 'x', in this case is just one. Thus the multiplication of these two terms leads to a result of '1.sin(x)'.

The product rule states that the newly ascertained expressions should be added together and therefore the derivative of 'y = x.sin(x)' is 'dy/dx = x.cos(x) + sin(x)'. Cool stuff, eh?

Answer:
**Parametric differentiation**

The method to solve this example of parametric differentiation starts by finding the derivatives of the two equations:

'x = cos^3 (t)'

'dx/dt = 3.cos^2 (t).(- sin(t))'

'y = sin^3 (t)'

'dy/dt = 3.sin^2 (t).(cos (t))'

To find 'dy/dx' from these two derivatives, simply divide 'dy/dt' by 'dx/dt' and this will simplify to '-tan(t)'. Therefore, by the differentiation of parametric equations, we have found that 'dy/dx = -tan(t)'.

(www.maths.abdn.ac.uk - source of this particular parametric differentiation example).

The method to solve this example of parametric differentiation starts by finding the derivatives of the two equations:

'x = cos^3 (t)'

'dx/dt = 3.cos^2 (t).(- sin(t))'

'y = sin^3 (t)'

'dy/dt = 3.sin^2 (t).(cos (t))'

To find 'dy/dx' from these two derivatives, simply divide 'dy/dt' by 'dx/dt' and this will simplify to '-tan(t)'. Therefore, by the differentiation of parametric equations, we have found that 'dy/dx = -tan(t)'.

(www.maths.abdn.ac.uk - source of this particular parametric differentiation example).

Answer:
**Implicit differentiation**

Implicit differentiation in this case would be differentiating 'y' when it should be the 'x' terms that are being differentiated. To differentiate the 'y' term in the equation '2y^2 = 3x^3', just use the usual methods. Thus, the derivative of this equation would be '4y = 9x^2', but that is not complete. After the 'y' term has been differentiated, a 'dy/dx' must be added. Ergo, the complete derivative is '4y.dy/dx = 9x^2'. Then the 'dy/dx' must be seen on its own so simply divide both sides by '4y' to get 'dy/dx = 9x^2/4y'.

In this example it wasn't essential to use implicit differentiation because the equation '2y^2 = 3x^3' could have been rearranged into the form 'y = ((3x^3)/2)^1/2'. However, that new form is a monstrosity and is not as simple to differentiate as when done implicitly.

Implicit differentiation in this case would be differentiating 'y' when it should be the 'x' terms that are being differentiated. To differentiate the 'y' term in the equation '2y^2 = 3x^3', just use the usual methods. Thus, the derivative of this equation would be '4y = 9x^2', but that is not complete. After the 'y' term has been differentiated, a 'dy/dx' must be added. Ergo, the complete derivative is '4y.dy/dx = 9x^2'. Then the 'dy/dx' must be seen on its own so simply divide both sides by '4y' to get 'dy/dx = 9x^2/4y'.

In this example it wasn't essential to use implicit differentiation because the equation '2y^2 = 3x^3' could have been rearranged into the form 'y = ((3x^3)/2)^1/2'. However, that new form is a monstrosity and is not as simple to differentiate as when done implicitly.

Answer:
**Integration**

One of the things that confuses people when they first start studying calculus is whether to integrate or differentiate in order to find the area under a curve. To integrate you just partake in the opposite process to differentiation. So if the equation of a curve was 'y = x^3 + 2x^2', then to integrate you would raise the power of each 'x' term by one and divide by the value of the new power. Thus, raising the power of 'x^3' by one would result in 'x^4' and to complete the integration of this term you divide it by four, '(x^4)/4'. Complete the same process for '2x^2' and you will get '(2x^3)/3'.

Whenever you integrate you must add in a constant, known simply as the constant of integration, which is usually given the letter 'C' in the UK. So, the area under the curve 'y = x^3 + 2x^2' could be found by adding values of 'x' into the fresh integration, '[(x^4)/4] + [(2x^3)/3] + C'.

One of the things that confuses people when they first start studying calculus is whether to integrate or differentiate in order to find the area under a curve. To integrate you just partake in the opposite process to differentiation. So if the equation of a curve was 'y = x^3 + 2x^2', then to integrate you would raise the power of each 'x' term by one and divide by the value of the new power. Thus, raising the power of 'x^3' by one would result in 'x^4' and to complete the integration of this term you divide it by four, '(x^4)/4'. Complete the same process for '2x^2' and you will get '(2x^3)/3'.

Whenever you integrate you must add in a constant, known simply as the constant of integration, which is usually given the letter 'C' in the UK. So, the area under the curve 'y = x^3 + 2x^2' could be found by adding values of 'x' into the fresh integration, '[(x^4)/4] + [(2x^3)/3] + C'.

This quiz was reviewed by FunTrivia editor crisw before going online.

Any errors found in FunTrivia content are routinely corrected through our feedback system.

Most Recent Scores

May 16 2023
:
Guest 129: 6/10May 16 2023 : Guest 38:

May 12 2023 : Guest 176: 1/10

May 12 2023 : TERRYHURST22:

Apr 22 2023 : Guest 76:

Apr 19 2023 : Guest 74: 8/10

Score Distribution

Related Quizzes

This quiz is part of series
1. **Complex Numbers: Real and Imaginary!** Average

2.**Interesting Indices in Incredible Instances!** Average

3.**Maths is Useless** Tough

4.**Circle Theorems** Average

5.**Straight Lines: The Knowledge** Average

6.**The Wonderful World of Differentiation** Average

7.**Questions on Quadratics** Average

2.

3.

4.

5.

6.

7.

Referenced Topics

World
People
Science
Vocabulary
Math
Names
The Internet
Mathematicians
Calculus
Other Destinations

Explore Other Quizzes by Go to

More

FunTrivia