Differentiation & Integration

Question 1

Use differentiation from first principles to find the gradient of the curve y=3x^2 at x=2

Answer 1

Let f(x)=3x^2, so f(x+h)=3(x+h)^2
=3x^2+6hx+3h^2
dy/dx=lim(h→0) (f(x+h)-f(x))/h
Substitute in values: (3x^2+6hx+3h^2-3x^2)/h = 6x+3h
The ‘h’ term vanishes as ‘h’ tends to zero so you’re left with dy/dx=6x, f(2)=6x2=12

Question 2

State the equation for the gradient function

Answer 2

dy/dx= lim(h→0) (f(x+h)-f(x))/h

Question 3

What is another way of saying dy/dx
?

Answer 3

f ’(x)

Question 4

What is the gradient function?

Answer 4

Formula for calculating a changing gradient

Question 5

Use differentiation from first principles to find the gradient of the curve y=x^2+2x

Answer 5

Let f(x)=x^2+2x, so f(x+h)=(x+h)^2+2(x+h) = x^2+2hx+h^2+2x+2h
Substitute into the equation: dy/dx= lim(h→0) (f(x+h)-f(x))/h
= (x^2+2hx+h^2+2x+2h - (x^2+2x))/h you must remember to multiply all of f(x) by -1 as it’s subtracted!!!
= (x^2+2hx+h^2+2x+2h-x^2-2x)/h
= 2x+h+2, the ‘h’ term vanishes as ‘h’ tends to zero so the answer is 2x+2

Question 6

What does f(x) in the gradient function formula represent?

Answer 6

The instantaneous gradient

Question 7

Differentiate this equation: y=x^5 + 3x^-2 + 5x^(1/2) + 10

Answer 7

Reduce each of the powers by 1 and multiply the co-efficient of x by the power: dy/dx = 5x^4 - 6x^-3 + 5/2x^-1/2 (+0)
The final term: 10 is an integer so differentiates to zero as y=10 would be a straight line with a gradient of zero

Question 8

What is the gradient function of e^x?

Answer 8

e^x because 1xe^x is e^x

Question 9

What does sin(x) differentiate into?

Answer 9

cos(x)

Question 10

What does cos(x) differentiate into?

Answer 10

-sin(x)

Question 11

Differentiate y= x^70 - x^-1 + 4x^7 + e^e^e^e + π^π^π

Answer 11

dy/dx= 70x^69 + x^-2 + 28x^6 (both e and π are representative of integers so differentiate to zero!)

Question 12

Differentiate y= 9x^-2/3

Answer 12

-6x^-5/3

Question 13

Differentiate y= x^π + e^π + e^2x

Answer 13

dy/dx= πx^π-1 + 0 + 2e^2x (e^π is an integer so differentiates to zero)
e^2x is differentiated using the first method we learnt (where you multiply the co-efficients), this is peculiar to ‘e’ being the base number

Question 14

Differentiate y= ((1/π)x)^2π +
(e^3)x^-e + e^πx

Answer 14

dy/dx= 2x^2π-1 - (e^4)x^-e-1 +
πe^ πx

Question 15

Given that:
f(x) = x^2+4x+12, x∈ℝ complete the square. Using this determine the largest value of 1/f(x)

Answer 15

(x+2)^2 +8
The largest value is when x=-2 (so that f(x)=(0)^2 +8, this means f(x)=8 so 1/f(x)=1/8

Question 16

Find the gradient of y=4-(2x^2) at the point (3,-14)

Answer 16

1) differentiate the equation by reducing the power: dy/dx=-4x
2) substitute in the x-coordinate: -4(3) =-12 (ignore the y coordinate you don’t need to use it!)

Question 17

Find the equation of the tangent and the normal to y=x^4 at the point (1,1)

Answer 17

1) dy/dx=4x^3
2) 4x(1)^3=4 so the gradient of the tangent is 4. y=mx+c 1=1x4+c, c=-3 so y=4x-3
3) gradient of the normal is the negative reciprocal (-1/4) y=mx+c 1=-1/4x1+c, c=5/4 y=-1/4x+5/4

Question 18

Why is the gradient function equal to the gradient of the tangent?

Answer 18

The gradient function equals the tangent’s gradient because differentiation finds the instantaneous rate of change (steepness) of a curve at a single point, and the tangent is defined as the straight line that perfectly touches the curve at that exact point, sharing the same steepness.

Question 19

Differentiate 1/(4x^3)

Answer 19

1) get it into a form you can differentiate: 1/(4x^3) = 1/4 x^-3
2) reduce the power: -3/4 x^-4

Question 20

Differentiate 2/(3√x)

Answer 20

1) get into a form you can differentiate: 2/3 x^-1/2
2) reduce the power: -1/3 x^-3/2

Question 21

Find f ’(x) for (x^2 - 3)/x

Answer 21

1) rearrange: x - 3/x = x - 3x^-1
2) differentiate: 1+3x^-2

Question 22

Why is 1/(2x^2) not equal to 2x^-2

Answer 22

2x^-2 = 2/x^2 not 1/(2x^2)

Question 23

A curve has equation:
y= x^3 +px^2+qx-45
The curve passes through the point R (2,3). The gradient of the curve at R is 8. Find the value of p and the value of q.

Answer 23

1) substitute the coordinates for R into the equation: 3= (2)^3+p(2)^2+ 2q-45, 40=4p+2q
2) differentiate the equation: dy/dx= 3x^2+2px+q
3) substitute x=2 and dy/dx=8 into the equation: 8= 3(2)^2+2p(2)+q, -4=4p+q
4) use simultaneous equations: p=-12, q=44

Question 24

Find the equation of the tangent to the curve y=e^3x when x=2

Answer 24

1) differentiate the equation: 3e^3x
2) substitute x=2: 3e^6 this is the gradient
3) find the y-coordinate by substituting x=2 into the equation y=e^3x to get y=e^6
4) substitute values into y=mx+c: e^6=(3e^6)x2+c, c=-5e^6
5) y=(3e^6)x - 5e^6

Question 25

What is the equation of the **tangent** to the curve y=x^3 at the point (-1,-1)?

Answer 25

y=mx+c, substitute in the values: -1=-m+c Work out the gradient using differentiation: dy/dx=3x^2 Dy/dx(-1)=3(-1)^2=3 so m=3 -1=-1x3+c, c=2 y=3x+2

Question 26

What is the equation of the **normal** to the curve: y=x^5 -5x that passes through the point (-2,-22)

Answer 26

Equation of the line is y=mx+c so -22=-2m+c Differentiate y=x^5 -5x = 5x^4 -5 Dy/dx(-2)=5(-2)^4 -5=75 The gradient of the normal is the negative reciprocal so 75 becomes -1/75 -22=(-2x-1/75)+c, c= -1652/75 y=-1/75 -1652/75

Question 27

Determine whether the curve 1/x^3 - x^3 is increasing or decreasing when x=1

Answer 27

Rearrange the equation to get rid of the quotient: y=x^-3 -x^3 dy/dx= -3x^-4 -3x^2 Rearrange to get -3/x^4 -3x^2 so it’s easier to deal with the substitution Substitute: dy/dx(1)= -3-3=-6 negative gradient means the curve is decreasing as dy/dx<0

Question 28

Find values of x for which the curve: y=x^3 + 3x^2 -5 is decreasing, write your answer in set notation

Answer 28

dy/dx= 3x^2 +6x dy/dx<0 when a curve is decreasing so 3x^2+6x<0 Factorise: 3x(x+2)<0 Roots are: x=0 and x=-2 (By drawing a sketch I know that when the curve is <0 the inequality is -2

Question 29

Describe the curve when dy/dx=0

Answer 29

The curve is flat (it’s a stationary point)

Question 30

Describe the curve when dy/dx>0

Answer 30

The curve is increasing

Question 31

Give the gradient function for a curve that’s decreasing

Answer 31

dy/dx<0

Question 32

When drawing the gradient function of a curve what should you start by doing?

Answer 32

Marking the turning points and drawing dashed lines to connect them to the x-axis (this will be where your line/ curve for dy/dx intersects the x-axis)

Question 33

When drawing gradient functions of curves if the gradient is negative where do you draw your dy/dx line. What about if the gradient is positive?

Answer 33

If it’s negative then start your line **below** the x-axis, if it’s positive start your line **above** the x-axis

Question 34

Are all stationary points turning points?

Answer 34

No (e.g. cubics that have 1 repeated root)

Question 35

What differentiates turning points from stationary points? What makes them the same.

Answer 35

The difference is that on either side of a turning point there’s a change in sign of the gradient, the similarity is that both mean dy/dx is instantaneously zero

Question 36

Find the stationary point of the graph y=x^3 How do you know it’s not a turning point?

Answer 36

dy/dx= 3x^2 dy/dx=0 3x^2=0 so x=0 There’s no change in sign of the gradient either side of x=0 both are positive. When x=1 you get 3, when x=-1 you get 3

Question 37

Find the stationary points of the graph y=x^3 -x^2 -x+4

Answer 37

dy/dx=3x^2 -2x-1 dy/dx=0 3x^2 -2x-1=0 (3x+1)(x-1)=0 x=-1/3 or x=1 Substitute into the original equation (y=x^3 -x^2 -x+4) so (-1/3,113/27) & (1,3) are the solutions

Question 38

What is the curvature function?

Answer 38

(d^2 y) / (dx^2)

Question 39

What are you doing if you set a curvature function to zero?

Answer 39

You are investigating for potential points of inflection

Question 40

What are points of inflection?

Answer 40

Points where there’s a curvature change (concave to convex or vice versa)

Question 41

How do you determine whether something is a point of inflection or not?

Answer 41

Set the **curvature function to zero** then **look for a sign change either side of the ‘x’** (sign change means curvature change)

Question 42

State the derivative of x^5

Answer 42

5x^4

Question 43

Where is the point of inflection on the graph of y=x^3 and is it stationary or non-stationary?

Answer 43

(0,0) the origin and it’s stationary (no change in sign of gradient)

Question 44

What conditions mean that a point is a stationary point of inflection?

Answer 44

dy/dx=0 and d^2y/dx^2=0 and there’s a change in sign either side of the value when using d^2y/dx^2 (curvature and gradient are instantaneouslyzero)

Question 45

What shape curvature means a curve is convex, what about concave?

Answer 45

Convex - ‘u’ shape Concave - ‘n’ shape

Question 46

What is the difference between stationary and non-stationary points of inflection?

Answer 46

Non-stationary points of inflection have a **gradient that’s not equal to zero**

Question 47

What are stationary points?

Answer 47

Points where the gradient (dy/dx) is instantaneously zero

Question 48

Does d^2y/dx^2=0 mean that a coordinate is a point of inflection?

Answer 48

Not necessarily, you must investigate either side of the point using d^2y/dx^2 looking for a change in sign

Question 49

What shape is a graph when d^2y/dx^2 > 0?

Answer 49

Convex (‘u’ shape)

Question 50

What shape is a graph when d^2y/dx^2 < 0?

Answer 50

Concave (‘n’ shape)

Question 51

What is the formula for the second derivative when an equation is equal to y?

Answer 51

d^2y/dx^2

Question 52

What is the formula for the second derivative when an equation is equal to f(x)?

Answer 52

f’’(x) **must put 2 dashes as 1 means the first derivative**

Question 53

Are local maximums and minimums points of inflection?

Answer 53

No, as there’s no change in curvature

Question 54

How do you investigate for local maximums and minimums?

Answer 54

1) Check dy/dx=0 2) Find d^2y/dx^2 3) If the answer is **>0 it’s a local minimum** 4) If the answer is **<0 it’s a local maximum** 5) **Or** instead of finding d^2y/dx^2 you can investigate either side of dy/dx for a change in sign

Question 55

Use differentiation to find the coordinates of the stationary point of the graph y=2+x -3x^2

Answer 55

dy/dx=-6x+1 Stationary points are when dy/dx=0 -6x+1=0, 6x=1, x=1/6 Insert into original equation of ‘y’ to find the y-coordinate (1/6,25/12)

Question 56

Optimisation: If A = 2x^2 + 864/x find the value of x for which A is a minimum, showing that it is a minimum

Answer 56

A = 2x^2 + 864x^-1 dA/dx = 4x - 864x^-2 **A minimum is a stationary point and stationary points are when dA/dx=0** so 4x - 864x^-2 =0 so x=6 d^2A/dx^2 = 4 + 1728x^-3 Insert x=6 into d^2A/dx^2 to get x=12. **A minimum is when d^2A/dx^2>0** and 12>0 so x=6 is a local minimum

Question 57

If A = x^2 + 1024/x find the value of x for which A is a minimum, showing that it is a minimum

Answer 57

A = x^2 + 1024x^-1 dA/dx = 2x - 1024x^-2 **A minimum is a stationary point and stationary points are when dA/dx=0** so 2x - 1024x^-2 =0 so x=8 d^2A/dx^2 = 2 + 2048x^-3 Insert x=8 into d^2A/dx^2 to get x=6. **A minimum is when d^2A/dx^2>0** and 6>0 so x=8 is a local minimum

Question 58

The coordinates of the stationary points of the curve: y = 2x^5 + 5x^4 are (0,0) and (-2,16), determine their types

Answer 58

(dy/dx = 10x^4 + 20x^3) d^2y/dx^2 = 40x^3 + 60x^2 d^2y/dx^2 (when x=-2) = -80<0 so (-2,16) is a local maximum d^2y/dx^2 (when x=0) = 0 so we don’t know, therefore have to try either side of dy/dx dy/dx (when x=-1) = -10<0 dy/dx (when x=1) = 30>0 The curve has a negative to positive gradient so is a local minimum (if it had no change in gradient it would be a stationary point inflection)

Question 59

What does determine their types mean in a differentiation question about coordinates?

Answer 59

Are the points local minima, maxima or points of inflection

Question 60

Determine the values of x for which the curve y = 3x^5 -10x^4 -50x^3 + 180x^2 is convex

Answer 60

dy/dx=15x^4 -40x^3 -150x^2 +360x d^2y/dx^2=60x^3 -120x^2 -300x +360 Divide by 60 = x^3 -2x^2 -5x +6 Factorise using polynomial division after testing different factors until you find a value of x that causes the polynomial to equal zero (often 1) When the curve is convex d^2y/dx^2>0 (x-1)(x+2)(x-3)>0 so x=1,-2,3 Sketch the cubic graph with those roots, when the curve is greater than zero the solution is: **-23**

Question 61

How do you solve the question: Determine the coordinates of any non-stationary points of inflection on the curve: y=x^4 + 2x^3 -12x^2 +2 ?

Answer 61

1) Find dy/dx 2) Find d^2y/dx^2 3) Points of inflection exist when d^2y/dx^2=0, this gives the potential points of inflection: (-2,-46) & (1,-7) 4) Insert the x values into dy/dx to check they aren’t stationary points, for -2 you get 40 and for -1 you get -14 neither of which equal zero 5) Use d^2y/dx^2 to check either side for sign changes (whether they’re points of inflection), (-2,-46) is convex to concave so yes and (1,-7) is concave to convex so yes both points satisfy the question

Question 62

Determine whether f(x)=x^3 -5x^2 +1 is increasing, decreasing or stationary when x=3

Answer 62

f ’(x)=3x^2 -10x f ‘(3)=3(3)^2 -10(3) = -3 < 0 so f(x) is decreasing when x=3

Question 63

Find the stationary points on the graph of y=x^3 -3x^2 -9x +5 and determine their types

Answer 63

dy/dx=3x^2 -6x -9 dy/dx=0 gives (x-3)(x+1)=0 so x=-1,3 (-1,10) and (3,-22) d^2y/dx^2=6x-6 When x=-1 6(-1)-6=-12<0 so it’s a local maximum When x=3 6(3)-6=12>0 so it’s a local minimum

Question 64

The maximum area of an enclosure is km^2. Find the value of k when A= 10x -2x^2

Answer 64

1) Find dy/dx and make it equal to zero (find the stationary point as this represents the maximum of the quadratic) dy/dx=10-4x, 10-4x=0, x=5/2 2) Insert x=5/2 into the equation for A to find its value: **A=12.5m^2**

Question 65

It is given that y=4√x, show that d^2y/dx^2 + (8/y^2)dy/dx =0

Answer 65

1) dy/dx= 2x^-1/2 2) d^2y/dx^2= -x^-3/2 3) substitute values: -x^-3/2 + (8/(4x^1/2)^2) x 2x^-1/2 = -x^-3/2 + x^-3/2 =0

Question 66

A circle has centre at the origin and radius R, this circle fits wholly inside the circle with equation: x^2 + y^2 -10x -24y =231. Determine the range of possible values of R

Answer 66

(x-5)^2 + (y-12)^2 =400, R=20 and c=(5,12) Distance from (5,12) to (0,0) is √(5^2)+(12^2) =13, 20-13=7 so R<7

Question 67

What are you actually doing to differentiate e^2x?

Answer 67

Differentiating the power to get 2, which you then multiply by e^2x to get 2e^2x

Question 68

Find the tangent line of y=2x^3 at (2,16)

Answer 68

dy/dx=6x^2 (same gradient as the line so simply substitute in the x-coordinate) dy/dx(2)=6(2)^2=24 16=24x2+c, c=32 y=24x+32

Question 69

The decrease in population of foxes over time is shown by the formula: P=230+180e^-0.05t. 1) state the initial population of foxes 2) find the rate at the which the population is decreasing when t=10

Answer 69

1) when t=0, 230+180=**410** 2) dy/dx=-9e^-0.05t, dy/dx(10) = **-5.459 per year**

Question 70

What does 1/x integrate to?

Answer 70

ln(x)

Question 71

What is integration used to find?

Answer 71

The area under a graph

Question 72

What is a particular differential equation?

Answer 72

A differential equation with no unknowns

Question 73

Integrate: dy/dx = x^3 -e^2x + 5x + 2

Answer 73

y= (x^4)/4 -(e^2x)/2 + (5x^2)/2 + 2x **+c**

Question 74

What are important things to remember when integrating?

Answer 74

1) always remember to add ‘c’ at the end!! 2) when integrating polynomials e.g. x^3 (regular values) **add 1 to the power, then divide by the new power** (doesn’t work when the power is -1 though) 3) when integrating with ‘e’ in the base, **differentiate the power, then divide by this number** e.g. for e^2x, 2x differentiated is 2 so the answer is (e^2x)/2

Question 75

When integrating e.g. dy/dx=x^2 what is the answer called?

Answer 75

y=(x^3)/3 +c is called the **anti-derivative**

Question 76

What does this symbol mean ∫ ?

Answer 76

It’s the **integration symbol and represents the summation** of infinitely many (rectangles) to get the area under a graph

Question 77

Find ∫ e^sin(x)

Answer 77

(e^sin(x))/cos(x) + c

Question 78

What are indefinite integrals?

Answer 78

Integrating **without bounds/ limits** (no numbers above and below the ∫ symbol), **you must add ‘c’** to these equations

Question 79

What are definite integrals?

Answer 79

Integrating **with bounds/ limits - the upper of which must be the larger number** (numbers above and below the ∫ symbol). **You don’t need to include ‘c’ in these equations**

Question 80

You are given that ∫(3x^-2 +4)dx Find y given that the curve passes through (1,5)

Answer 80

1) y=(3x^-1)/-1 + 4x +c 2) simplify: y=-3/x + 4x +c 3) substitute in the coordinate (1,5): 5=-3/1 + (4x1) + c so c=4 4) y= -3/x +4x +4

Question 81

What is another way of writing: dy/dx = 3/x^2 +4 as an integration?

Answer 81

∫(3x^-2 +4)dx

Question 82

Find the area ∫(x^3)dx when the upper bound is 10 and the lower bound is 7

Answer 82

1) integrate: (x^4)/4 2) substitute in x=10 and x=7 to the equation to get (10^4)/4 =2,500 and (7^4)/4=600.25 3) **subtract the lower bound substitution from the upper bound** to get 2500-600.25 =1,899.75 **units^2**

Question 83

What should you do if you get a negative area when integrating?

Answer 83

**Take the additive inverse** (it doesn’t mean you’re wrong if you get a negative area it just means the area is below the x-axis)

Question 84

Find the gradient of the tangent to the curve y=x^1/2 at the point x=4

Answer 84

To find the gradient of the curve find dy/dx = 1/2x^-1/2. To find the gradient of the tangent, substitute the point x=4 into dy/dx = 1/2x^-1/2. To get 1/4

Question 85

How can you simplify ∫(1/2x)dx

Answer 85

∫(1/2 x1/x) dx = **1/2 ∫**(1/x)dx = 1/2ln(x) + c When you have one multiplier of x you can take it outside the integration function and simply multiply your final answer by that value (in this case 1/2)

Question 86

How is ∫(ax^n)dx simplified?

Answer 86

a ∫(x^n)dx the ‘a’ multiplier can be removed from the integration function and you can multiply your final answer by the ‘a’

Question 87

How can you simplify ∫(-x^2)dx with an upper bound of ‘a’ and a lower bound of ‘b’?

Answer 87

1) **swap the bounds**! So ‘a’ becomes the lower bound and ‘b’ the upper bound 2) **change the sign in the brackets** 3) final answer: ∫(x^2)dx with lower bound ‘a’ and upper bound ‘b’

Question 88

What should you do if a graph intersects the x-axis in between two limits when you’re integrating?

Answer 88

**Add together the values you get for the upper and lower bound** (rather than subtracting the lower from the upper) otherwise you will get an incorrect answer of zero!

Question 89

Find ∫(x^3)dx with the upper bound 1 and lower bound -1

Answer 89

1) integrate: (x^4)/4 2) when x=1 you get 1/4 3) when x=-1 you get 1/4 4) **add the answers** otherwise you get an incorrect answer of zero (because the graph crosses the x-axis)! 5) final answer: **1/2 units^2**

Question 90

T=22+50e^-1/8t. Determine the value of T when the drink is cooling at a rate of 2.5 degrees Celsius per minute

Answer 90

dT/dt = -25/4 e^-1/8t so t=-8ln(2/5) T=22+50x2/5=42 so T=42 degrees

Question 91

Find y as a function of x for this equation: dy/dx=5x^4 +3

Answer 91

Integrate: y=x^5 + 3x **+c**

Question 92

Evaluate: ∫ x(x^2 -4)dx with the upper bound 2 and the lower bound -2. Explain the result.

Answer 92

Expand the bracket: x^3 -4x Integrate: (x^4)/4 -2x^2 When x=2: 4-8 When x=-2: 4-8 (4-8)-(4-8)=0 as the areas are equal and the integrals have opposite signs I=0

Question 93

Differentiate from first principles: y=4x^2 +x

Answer 93

1) **find f(x+h) when f(x)=4x^2 +x:** f(x+h)=4(x+h)^2 + (x+h) = 4x^2 +8hx + 4h^2 +x +h 2) **find the difference in y values:** f(x+h)-f(x) = 4x^2 +8hx + 4h^2 +x +h - (4x^2 +x) = 8hx + 4h^2 +h 3) **find the gradient (change in y/ change in x):** (8hx + 4h^2 +h) / (x+h)-x = 8x + 4h +1 4) **as h tends to zero:** you get the answer **8x+1**