Differentiation past exam notes

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Resource booklet

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2017 Question one

(a) Differentiate y = (x)^1/2 + tan(2x).

Achievement question

dy/dx=1/2.x^-1/2 + 2sec^2 (2x)

(b) Find the gradient of the tangent to the curve y = e^2x/(x + 2) at the point where x = 0.

Achievement question

use Quotent rule

dy/dx= ((x+2)2e^2x-e^2x)/(x+2)^2

when X=0

dy/dx=3/4

(c) The normal to the parabola y = 0.5(x – 3)^2 + 2 at the point (1,4) intersects the parabola again at the point P.

Find the x-coordinate of point P.

merit question

dy/dx=2*1/2(x-3)=x-3

when x=1

dy/dx=-2

therefore the gradient of the tangent at (1,4) is -2

so the gradient of the normal m=1/2

the equation of the normal, y=1/2x+c passes through (1,4) so 4=1/2*1+c

c=7/2

y=1/2x+7/2

At point P 1/2x+/7/2=0.5(x-3)^2 +2

x+7=(x-3)^2 +4

x+7=x^2-6x+13

x^2-7X+6=0

(x-6)(x-1)=0

at point P, x=6

(d) A curve is defined parametrically by the equations x = (t +1)^1/2 and y = sin (2t).

Find the gradient of the tangent to the curve at the point when t = 0.

Merit question

dx/dt=1/2(t+1)^-1/2

dy/dt=2.cos(2t)

dy/dx=dy/dt.dt/dx

dy/dx=2.cos(2t).2(t+1)^1/2

when t=0, dy/dx=4

(e) Find the values of a and b such that the curve y = (ax − b)/(x^2 −1) has a turning point at (3,1)

Excellence question

dy/dx=a.(x^2-1)-2x.(xa-b)/(x^2-1)^2

at x=3 and dy/dx=0

8a-6(3a-b)=0

-10a+6b=0

5a=3b

the curve passes through (3,1)

1=(3a-b)/8

(8=3a-b)*3

24=9a-3b (where 3b=5a)

24=9a-5a

24=4a

a=6

b=10

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2017 Question two

(a) Differentiate y = 2(x^2 – 4x)^5.

acheived question

dy/dx=10(x^2-4x)^4.(2x-4)

(b) The percentage of seeds germinating depends on the amount of water applied to the seedbed that the seeds are sown in, and may be modeled by the function:

P(w) = 96ln(w + 1.25) – 16w – 12

where P is the percentage of seeds that germinate and w is the daily amount of water applied (liters per square meter of seedbed), with 0 ≤ w ≤ 15. Find the amount of water that should be applied daily to maximize the percentage of seeds germinating.

Achieved question

p’(w)=96/(w+1.25)-16

Maximum when p’(w)=0

0=96/(w+1.25)-16

16(w+1.25)=96

16w+20=96

16w=76

w=4.75

(c) The tangent to the curve y = √x is drawn at the point (4,2).

Find the coordinates of the point Q where the tangent intersects the x-axis.

Merit question

y=√x

dy/dx=1/2x-½

=1/2√x

At (4,2) dy/dx=1/2√4=1/4

tangent: y=1/4x+c through (4,2)

2=1+c, c=1

y=1/4x+1

when y=0 , 0=1/4x+1

x=-4

Point Q is at (-4,0)

(d) Find the coordinates of the point P (x,y) on the curve y =√x that is closest to the point (4,0).

You do not need to prove that your solution is the minimum value

Merit question

d²=(4-x)²+(√x)²

=16-8x+x²+x

=x²-7x+16

d(d²)/dx=2x-7

at minimum d(d²)/dx=0

0=2x-7

x=3.5

y=√x, y=√3.5

p=(3.5,√3.5)

(e) A rectangle is inscribed in a semi-circle of radius r, as shown below.

Show that the maximum possible area of such a rectangle occurs when x = r/√ 2.

Excellence question

Area= 2x√(r²-x²)

A(x)=2x.(r²-x²)½

A’(x)=x.(r²-x²)½ +2x.(r²-x²)–½ ×1/2×-2x

=2.(r²-x²)½ +(-2x²).(r²-x²)–½

=2√(r²-x²)-2x²/√(r²-x²)=0

2√(r²-x²)=2x²/√(r²-x²)

√(r²-x²)=x²/√(r²-x²)

x²=(r²-x²)

2x²=r²

x²=r²/2

x=r/√2

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2017 Question three

(a) Differentiate y = x .ln(3x – 1)

Achieved question

dy/dx=ln(3x-1)+x/(3x-1).3

(b) Find the gradient of the curve y=1/x-1/x² at the point (2,1/4)

Merit question

y=1/x-1/x²

y=x-1 -x-2

dy/dx=-1/x²+2/x³

At x=2 dy/dx=-1/2²+2/2³

dy/dx=0

(c) The graph below shows the function y = f(x).

For the function above:

(i) Find the value(s) of x that meet the following conditions:

Merit question

(1) f'(x) = 0: x<-2, x=2

(2) f(x) is continuous but not differentiable: -2, 1

(3) f(x) is not continuous: -1, 0

(4) f''(x) < 0: x>1

(ii) What is the value of x→ −1 lim f (x)?

State clearly if the value does not exist.

x→ −1 lim f (x) =2

(d) A building has an external elevator. The elevator is rising at a constant rate of 2 ms–1.

Sarah is stationary, watching the elevator from a point 30 m away from the base of the elevator shaft.

Let the angle of elevation of the elevator floor from Sarah's eye level be θ

Find the rate at which the angle of elevation is increasing when the elevator floor is 20 m above Sarah’s eye level.​​​​​​​

Merit question

let h=height above sarah’s eye level

tanθ=(h/30)

h=30tanθ

dh/dθ=30sec²θ

dh/dt=2

dθ/dt=dθ/dh×dh/dt

dθ/dt= 2×1/30sec²θ

=cos²θ/15

at h=20 θ=tan-1 (20/30)=0.588...

=(cos(5.88))²/15

=0.046 rad/s

(e) For the function y = ex .cos(kx):

(i) Find dy/dx and d²y/dx².

Merit question

dy/dx=ex .cos(kx)+ex .(-ksin(kx))

=ex (cos(kx)-ksin(kx))

d²y/dx²=ex (cos(kx)-ksin(kx))+ex (-ksin(kx)-k²cos(kx))

=ex (cos(kx)-2ksin(kx)-k²cos(kx))

(ii) Find all the value(s) of k such that the function y = excos(kx) satisfies the equation d²y/dx²−2dy/dx+2y=0.

for all values of x.

Excellence question

dy/dx=ex (cos(kx)-ksin(kx))

d²y/dx²=ex (cos(kx)-2ksin(kx)-k²cos(kx))

ex (cos(kx)-2ksin(kx)-k²cos(kx))-2(ex (cos(kx)-ksin(kx))+2(excos(kx))=0

ex (cos(kx)-k²cos(kx))=0

ex cos(kx) (1-k²)=0

k=±1

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2017 Exam summary

Question one

achieved

find the derivative of a function

find the gradient of the tangent

Merit

find the equation of a normal line, and use to locate the coordinates of a point

use the parametric equation to find the gradient of the tangent

form simultaneous equations relating two unknowns

Excellence

use simultaneous equations to find unknowns of a function

Question two

Achieved

find the derivative of a composite function using the chain rule

solve a maximization application of a function

find the derivative of a function*

find the second derivative of a function

Merit

find the equation of a tangent, and use to find the coordinates of a point

form an equation to find the minimum distance of a point and find the coordinates to that point

form an equation to find the maximum area in an application situation and differentiate

Excellence

use formed equation to find a maximum area

Question three

Achieved

find the derivative of a function*

find the derivative of a product using the product rule

find gradient at a point

graph features including continuity, differentiability

Merit

graph features including values of x when f('x)=0, values of x when f''(x)<0

limits

related rates of change using the parametric function

find the second derivative of a product function

Excellence

substitute function derived function and 2nd derived function into an equation to solve for an unknown.

Summary

Achieved

find the derivative of a function

find the second derivative of a function

find gradient at a point

find the gradient of the tangent

find the derivative of a composite function using the chain rule

find the derivative of a product using the product rule

identify graph features including continuity, differentiability

solve a maximization application of a function

Merit

differentiate a parametric equation to find the gradient of the tangent

find the equation of a tangent, and use to find the coordinates of a point

find the equation of a normal line, and use to locate the coordinates of a point

form simultaneous equations relating two unknowns

form an equation to find the minimum distance of a point and find the coordinates to that point

form an equation to find the maximum area in an application situation and differentiate

identify graph features including values of x when f('x)=0, values of x when f''(x)<0

limits

related rates of change using the parametric function

find the second derivative of a product function

Excellence

use simultaneous equations to find unknowns of a function

use formed equation to find a maximum area

substitute function derived function and 2nd derived function into an equation to solve for an unknown.

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2016