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

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Quiz Answer Key and Fun Facts

Answer:
**Tangent**

The word tangent may be familiar from its use to name one of the three trigonometric functions commonly met in high school (along with the sine and the cosine). In geometry, it means a line or curve which touches another one at exactly one point. The word comes from the Latin verb 'tangere', to touch.

The word tangent may be familiar from its use to name one of the three trigonometric functions commonly met in high school (along with the sine and the cosine). In geometry, it means a line or curve which touches another one at exactly one point. The word comes from the Latin verb 'tangere', to touch.

Answer:
**First derivative of the curve**

The first derivative is a function of the independent variable (usually x in the examples to follow), which is found by applying calculus. For example, the derivative of the curve y = x^2 (read y equals x squared) is dy/dx = 2x. The value of this derivative may change at various points along the curve, so to find how steep the curve is at any point, you insert the x-coordinate of that point into the equation for the derivative, and evaluate the result. For the example, at the point where x = 3, the gradient is 2*3, or 6.

This means that the curve is going up as steeply as a line of gradient 6.

The first derivative is a function of the independent variable (usually x in the examples to follow), which is found by applying calculus. For example, the derivative of the curve y = x^2 (read y equals x squared) is dy/dx = 2x. The value of this derivative may change at various points along the curve, so to find how steep the curve is at any point, you insert the x-coordinate of that point into the equation for the derivative, and evaluate the result. For the example, at the point where x = 3, the gradient is 2*3, or 6.

This means that the curve is going up as steeply as a line of gradient 6.

Answer:
**Isaac Newton**

All four of these, along with Isaac Barrow, are considered the founders of calculus. As is often the case, it is not clear exactly who came up with which idea when (there was no internet to allow virtually instantaneous publication and dispersal of ideas), but many introductory calculus texts refer to the use of Newton's Method.

It involves writing the equation of a line which goes from a specified point on your curve (x0) to any other point, and seeing what happens as the other point is moved closer and closer to the fixed point.

The technical phrase is that we find the limit as x approaches x0. Don't worry, we're not going to actually do it!

All four of these, along with Isaac Barrow, are considered the founders of calculus. As is often the case, it is not clear exactly who came up with which idea when (there was no internet to allow virtually instantaneous publication and dispersal of ideas), but many introductory calculus texts refer to the use of Newton's Method.

It involves writing the equation of a line which goes from a specified point on your curve (x0) to any other point, and seeing what happens as the other point is moved closer and closer to the fixed point.

The technical phrase is that we find the limit as x approaches x0. Don't worry, we're not going to actually do it!

Answer:
**The curve is horizontal at that point**

A positive gradient means the curve is increasing in value as x increases, and the curve is going up; a negative value means it is going down. A vertical line does not have a gradient, while a horizontal one has a gradient of zero. This fact can be used to find the point where a curve has a maximum or minimum value - if it changes from having a positive gradient to having a negative one, then the point where the derivative has a value of 0 is a local maximum value; if it is changing from a negative value to a positive value, the point is a local minimum.

A positive gradient means the curve is increasing in value as x increases, and the curve is going up; a negative value means it is going down. A vertical line does not have a gradient, while a horizontal one has a gradient of zero. This fact can be used to find the point where a curve has a maximum or minimum value - if it changes from having a positive gradient to having a negative one, then the point where the derivative has a value of 0 is a local maximum value; if it is changing from a negative value to a positive value, the point is a local minimum.

Answer:
**20x^3**

The derivative of x^n, using the rule, would be 4x^3. Then you have to multiply by the coefficient of 5. 5x4=20, so the result is 20x^3. Using this rule, you can show that the derivative of a linear function is just a number, which is the unchanging gradient of the line.

Another interesting and useful result is that the derivative of a function such as y=4, which can be drawn as a horizontal line, is 0.

The derivative of x^n, using the rule, would be 4x^3. Then you have to multiply by the coefficient of 5. 5x4=20, so the result is 20x^3. Using this rule, you can show that the derivative of a linear function is just a number, which is the unchanging gradient of the line.

Another interesting and useful result is that the derivative of a function such as y=4, which can be drawn as a horizontal line, is 0.

Answer:
**6x + 5**

As in the previous question, the derivative of 3x^2 is 2*3x^1, or 6x. The derivative of 5x is 1*5x^0. Since x^0 is 1 (as is the case for any number to the power of zero), this can just be written as 5. No matter how many terms you are adding together, and no matter how complicated they may look, you can always just differentiate them one at a time, and add the results together. One tricky thing - if the derivative turns out to have a negative sign in front of it, you are adding a negative term, which is the same as subtracting it.

As in the previous question, the derivative of 3x^2 is 2*3x^1, or 6x. The derivative of 5x is 1*5x^0. Since x^0 is 1 (as is the case for any number to the power of zero), this can just be written as 5. No matter how many terms you are adding together, and no matter how complicated they may look, you can always just differentiate them one at a time, and add the results together. One tricky thing - if the derivative turns out to have a negative sign in front of it, you are adding a negative term, which is the same as subtracting it.

Answer:
**Product Rule**

The product of two expressions means they have been multiplied together, and the Product Rule tells you how to find the derivative of a product. The Quotient Rule can be used when you have one expression being divided by another; the Chain Rule is used when you have multiple processes being applied (we'll look at this in a minute), and the term Sum Rule is not usually used, but would apply to a polynomial.

If you want to know how the product rule works, here goes. The letters u and v describe the two expressions in x that are being multiplied together. If y = uv, then dy/dx = u*dv/dx + v*dy/dx.

For the example in the question, we let u = x^3+2, and v = 3x-1.

Differentiating each of these, du/dx = 3x^2+2 and dv/dx = 3.

Substituting into the rule, dy/dx = (x^3+2x)(3) + (3x-1)(3x^2+2).

Next step is to expand each term, giving dy/dx = 3x^3 + 6x + 9x^3 +6x -3x^2 - 2.

Now collect up like terms, and write them in order of decreasing powers of x: dy/dx = 12x^3 - 3x^2 + 12x -2.

The product of two expressions means they have been multiplied together, and the Product Rule tells you how to find the derivative of a product. The Quotient Rule can be used when you have one expression being divided by another; the Chain Rule is used when you have multiple processes being applied (we'll look at this in a minute), and the term Sum Rule is not usually used, but would apply to a polynomial.

If you want to know how the product rule works, here goes. The letters u and v describe the two expressions in x that are being multiplied together. If y = uv, then dy/dx = u*dv/dx + v*dy/dx.

For the example in the question, we let u = x^3+2, and v = 3x-1.

Differentiating each of these, du/dx = 3x^2+2 and dv/dx = 3.

Substituting into the rule, dy/dx = (x^3+2x)(3) + (3x-1)(3x^2+2).

Next step is to expand each term, giving dy/dx = 3x^3 + 6x + 9x^3 +6x -3x^2 - 2.

Now collect up like terms, and write them in order of decreasing powers of x: dy/dx = 12x^3 - 3x^2 + 12x -2.

Answer:
**y = z^2**

The letter z represents the first step, doing something to x, so in the example z = 5x^4 + 4x^3 - 7. (Remember the Order of Operations rule, which says the first thing to be calculated is what is inside the brackets?) The original expression says that we now have to square the result of that process, so y = z^2. If that is enough information for you, move on to the next question. If you want to see how the rule works in this case, read the next paragraph.

To use the chain rule, we need to find dy/dz and dz/dx. Since y = z^2, dy/dz = 2z.

To write this in terms of x, we use the fact that z = 5x^4 + 4x^3 - 7, which means that 2z = 10x^4 + 8x^3 - 14.

Since z = 5x^4 + 4x^3 - 7, dz/dx = 20x^3 + 12x^2.

Their product is (10x^4 + 8x^3 - 14) (20x^3 + 12x^2).

Next step is to expand, giving 200x^7 + 160x^6 -280x^3 + 120x^6 + 72x^5 - 168x^2.

Almost there, collect the like terms and write them in order of decreasing powers of x: 200x^7 + 280x^6 + 72x^5 - 280x^3 - 168x^2.

Simple, really. Or not.

The letter z represents the first step, doing something to x, so in the example z = 5x^4 + 4x^3 - 7. (Remember the Order of Operations rule, which says the first thing to be calculated is what is inside the brackets?) The original expression says that we now have to square the result of that process, so y = z^2. If that is enough information for you, move on to the next question. If you want to see how the rule works in this case, read the next paragraph.

To use the chain rule, we need to find dy/dz and dz/dx. Since y = z^2, dy/dz = 2z.

To write this in terms of x, we use the fact that z = 5x^4 + 4x^3 - 7, which means that 2z = 10x^4 + 8x^3 - 14.

Since z = 5x^4 + 4x^3 - 7, dz/dx = 20x^3 + 12x^2.

Their product is (10x^4 + 8x^3 - 14) (20x^3 + 12x^2).

Next step is to expand, giving 200x^7 + 160x^6 -280x^3 + 120x^6 + 72x^5 - 168x^2.

Almost there, collect the like terms and write them in order of decreasing powers of x: 200x^7 + 280x^6 + 72x^5 - 280x^3 - 168x^2.

Simple, really. Or not.

Answer:
**Acceleration**

The displacement is found using the original function describing its position. The velocity is measured by the first derivative with respect to time (t) of the position function, and the acceleration, which measures how quickly velocity is changing, by the second derivative. If you want to see whether the acceleration is changing over time, which is called the surge, you need to look at the third derivative. To find the second derivative, you just find the derivative of the first derivative. For example, if position = 4t^3, then velocity = 12t^2, and acceleration is the derivative of 12t^2, which is 24t.

The displacement is found using the original function describing its position. The velocity is measured by the first derivative with respect to time (t) of the position function, and the acceleration, which measures how quickly velocity is changing, by the second derivative. If you want to see whether the acceleration is changing over time, which is called the surge, you need to look at the third derivative. To find the second derivative, you just find the derivative of the first derivative. For example, if position = 4t^3, then velocity = 12t^2, and acceleration is the derivative of 12t^2, which is 24t.

Answer:
**Antidifferentiation**

As an example, we might know the velocity of an object, and wish to find where it will be at some time in the future. Antidifferentiating the velocity rule will give us the position rule. In order to work out exactly where it will be, we need to know where it was at the start (or at some other time), and then can calculate its final position. If the velocity is given as 4t+3, and the object started at position 10, then its future position can be given by the rule d = 2t^2 + 3t + 10. (Can you see how this rule for d can be differentiated to give the original rule for v?) Almost any function can be differentiated, with enough care and effort, but antidifferentiation is much more complex.

Many functions must be antidifferentiated using tables of rules produced over the centuries by differentiating complicated functions, and noting the result.

As an example, we might know the velocity of an object, and wish to find where it will be at some time in the future. Antidifferentiating the velocity rule will give us the position rule. In order to work out exactly where it will be, we need to know where it was at the start (or at some other time), and then can calculate its final position. If the velocity is given as 4t+3, and the object started at position 10, then its future position can be given by the rule d = 2t^2 + 3t + 10. (Can you see how this rule for d can be differentiated to give the original rule for v?) Almost any function can be differentiated, with enough care and effort, but antidifferentiation is much more complex.

Many functions must be antidifferentiated using tables of rules produced over the centuries by differentiating complicated functions, and noting the result.

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

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

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