FREE! Click here to Join FunTrivia. Thousands of games, quizzes, and lots more!
Quiz about Maths Topic 7 Introductory Calculus
Quiz about Maths Topic 7 Introductory Calculus

Maths Topic 7: Introductory Calculus Quiz


This quiz will cover basic differentiation rules and equations of tangents and normals. You will need a pen, paper and a calculator for this quiz. Good luck.

A multiple-choice quiz by dialga483. Estimated time: 5 mins.
  1. Home
  2. »
  3. Quizzes
  4. »
  5. Science Trivia
  6. »
  7. Math
  8. »
  9. Math Problems

Author
dialga483
Time
5 mins
Type
Multiple Choice
Quiz #
366,212
Updated
Dec 03 21
# Qns
10
Difficulty
Tough
Avg Score
6 / 10
Plays
227
Last 3 plays: Guest 12 (3/10), Melodia19 (7/10), Guest 50 (1/10).
-
Question 1 of 10
1. The gradient of the secant to the curve y=f(x), passing through the points (x,f(x)) and (a,f(a)) is given by which expression? Hint


Question 2 of 10
2. The differentiation of a function by first principles is given by which expression? Hint


Question 3 of 10
3. What is the gradient of the tangent to the curve y=3x^2 at the point x=2? Hint


Question 4 of 10
4. What is the derivative of the function f(x)=x^2-4x-3? Hint


Question 5 of 10
5. What is the derivative of the function f(x)=sqrt(x)? Hint


Question 6 of 10
6. What is the derivative of the function f(x)=(2x+7)^5? Hint


Question 7 of 10
7. Using the product rule, what is the derivative of the function f(x)=2x^5 (5x+7)^3? Hint


Question 8 of 10
8. Using the quotient rule, what is the derivative of the function
f(x)=(3x-5)/(5x+2)?
Hint


Question 9 of 10
9. What is the equation of the tangent to the curve y=2x^3-7 at the point where x=1? Hint


Question 10 of 10
10. What is the equation of the normal to the curve y=(2x+3)/(5x-1) at the point
(-1,(-1/6))?
Hint



(Optional) Create a Free FunTrivia ID to save the points you are about to earn:

arrow Select a User ID:
arrow Choose a Password:
arrow Your Email:




Most Recent Scores
Mar 12 2024 : Guest 12: 3/10
Feb 01 2024 : Melodia19: 7/10
Jan 30 2024 : Guest 50: 1/10

Score Distribution

quiz
Quiz Answer Key and Fun Facts
1. The gradient of the secant to the curve y=f(x), passing through the points (x,f(x)) and (a,f(a)) is given by which expression?

Answer: m=(f(x)-f(a))/(x-a)

The secant to a curve can be defined as the straight line passing through two points on the curve. Therefore, we can work out the gradient in the same way we work out gradient of any straight line, by using the rule m=(y2-y1)/(x2-x1).
2. The differentiation of a function by first principles is given by which expression?

Answer: f'(x)=lim h->0 (f(x+h)-f(x))/h

To solve this, we can consider two points on a curve, (x,f(x)) and (x+h,f(x+h)). Since the derivative can be used to find the gradient of a tangent, we can use the two points to show that f'(x)=lim h->0 (f(x+h)-f(x))/h.
3. What is the gradient of the tangent to the curve y=3x^2 at the point x=2?

Answer: 12

To solve this question, we simply need to differentiate the curve.
So,
y=3x^2
dy/dx=6x
Now at x=2
dy/dx=6(2)
=12
4. What is the derivative of the function f(x)=x^2-4x-3?

Answer: f'(x)=2x-4

To differentiate a polynomial function, we simply differentiate each term separately
So,
f(x)=x^2-4x-3
f'(x)=2x-4
5. What is the derivative of the function f(x)=sqrt(x)?

Answer: 1/(2sqrt(x))

To differentiate this function, we first need to re-write it in index form
So
sqrt(x)=x^(1/2)
Now differentiating this gives,
f'(x)=(1/2)x^(-1/2)
f'(x)=(1/2)(1/x^(1/2))
f'(x)=1/(2sqrt(x))
6. What is the derivative of the function f(x)=(2x+7)^5?

Answer: f'(x)=10(2x+7)^4

To differentiate this function, we need to use the rule "if y=(f(x))^n, then dy/dx = n(f(x))^(n-1) * f'(x)"
So if f(x)=(2x+7)^5
Then,
f'(x)=5(2x+7)^4 * 2
f'(x)=10(2x+7)^4
7. Using the product rule, what is the derivative of the function f(x)=2x^5 (5x+7)^3?

Answer: f'(x)=(10x^4) ((5x+7)^2) (8x+7)

The product rule states that "if y=f(x)g(x), then dy/dx=f'(x)g(x)+g'(x)f(x)"
Now if we apply this to the given function,
f'(x)=(10x^4(5x+7)^3)+(6x^5(5x+7)^2*5)
f'(x)=(10x^4(5x+7)^3)+(30x^5(5x+7)^2)
Now simplifying this,
f'(x)=(10x^4)((5x+7)^2)((5x+7)+3x)
f'(x)=(10x^4)((5x+7)^2)(8x+7)
8. Using the quotient rule, what is the derivative of the function f(x)=(3x-5)/(5x+2)?

Answer: f'(x)=31/(5x+2)^2

The quotient rule states that
"if y=f(x)/g(x), then dy/dx=(f'(x)g(x)-g'(x)f(x))/(g(x))^2
Now if we apply this to the given function
f'(x)=(3(5x+2)-5(3x-5))/(5x+2)^2
f'(x)=(15x+6-15x+25)/(5x+2)^2
f'(x)=31/(5x+2)^2
9. What is the equation of the tangent to the curve y=2x^3-7 at the point where x=1?

Answer: 6x-y-11=0

To solve this question, we first need to differentiate the curve to find the gradient
So,
y=2x^3-7
dy/dx=6x^2
at x=1
dy/dx=6
Now we need to find our y-coordinate
So, at x=1
y=2(1)^3-7
y=-5
Now using the rule y-y1=m(x-x1)
y+5=6(x-1)
y+5=6x-6
6x-y-11=0
Therefore, the equation of the tangent is 6x-y-11=0.
10. What is the equation of the normal to the curve y=(2x+3)/(5x-1) at the point (-1,(-1/6))?

Answer: 216x-102y+199=0

To solve this question, we first need to differentiate the curve to find the gradient.
So,
y=(2x+3)/(5x-1)
dy/dx=(2(5x-1)-5(2x+3))/(5x-1)^2
dy/dx=(10x-2-10x-15)/(5x-1)^2
dy/dx=-17/(5x-1)^2
at x=-1
dy/dx=-17/(5(-1)-1)^2
dy/dx=-17/36
Since this is the gradient of the tangent, we need to take the reciprocal to find the gradient of the normal
Therefore, dy/dx=36/17
Now using the rule y-y1=m(x-x1)
y+(1/6)=(36/17)(x+1)
17y+(17/6)=36(x+1)
17y+(17/6)=36x+36
102y+17=216x+216
216x-102y+199=0
Therefore, the equation of the normal is 216x-102y+199=0.
Source: Author dialga483

This quiz was reviewed by FunTrivia editor WesleyCrusher before going online.
Any errors found in FunTrivia content are routinely corrected through our feedback system.
3/28/2024, Copyright 2024 FunTrivia, Inc. - Report an Error / Contact Us