The 1st and 2nd Derivatives & Applications of Derivatives Flashcards
What is f’(x)
Instantaneous rate of change - The gradient of the tangent to the curve at a point
if f’(x) >0
gradient is increasing
if f’(x) <0
gradient is decreasing
if f’(x) =0
gradient is stationary
Stationary point
gradient of the tangent to the curve at the point is horizontal
Maximum turning point
gradient goes from-ve –>0–>+ve
i.e /-\
Minimum turning point
gradient goes from +ve –> 0 –> +ve
i.e _/
point of inflection
point on curve where tangent crosses the line
i.e the concavity changes
local maximum
Given point (a, f(a)) if f(x) is less than or equal to f(a)
local minimum
given the point (a, f(a)) if f(x) is greater than or equal to f(a)
how to find stationary points
Let the first derivative =0
global maximum
the highest point at the endpoint of a given domain
global minimum
lowest point at the other endpoint of a given domain
how to find max and min value of a function in a given domain
sub each value of x in the domain into the function
Second derivative
the rate of change of the first derivative (the rate of change of the gradient)
if f’‘(a)>0
concave upwards and there is a minimum turning point there
if f’‘(a)<0
concave downwards and there is a maximum turning point there
if y’‘=0
there is an inflection point at a point on the curve AND concavity changes at this point
max turning point…
y’=0 and y’‘<0
min turning point…
y’=0 and y’‘>0
horizontal point of inflection…
y’=0, y’‘=0 and concavity changes
when y’‘(a)=0
more work is needed to decide the nature of the point
what table to use with f(x)
table of signs –> to find where the function is positive or negative
what table to use with f’(x)
table of slopes –> to determine the nature of stationary points
what table to use with f’‘(x)
table of concavity –> to find any points of inflection and the concavity of the function
point of inflection
concavity changes
how to find stationary points
let y’=0
how to find global max and min
examine and compare the: turning points, boundaries of the domain (or behaviour for large x) and any discontinuities of f’(x) (when y=0/0)
steps to find the nature of stationary points
derive, solve y’=0
if stationary points exist –> sub into original to find coords
easy to differentiate–> second derive –> sign of y’’
- zero –> table of concavities
- positive –> min turning point
- negative –> max turning point
not easy to differentiate –> table of slopes with y’
_/ –> min turning point
/-\ –> max turning point
— –> horizontal point of inflection
how to solve max/min qs
the three cs
construct: draw a diagram, assign pronumerals/form equations, note restrictions
calculus: differentiate, find T.Ps then find nature
conclude: evaluate and compare T.Ps with endpoints/discontinuities, state final answer
Displacement
Relative position to starting point
when displacement is negative
particle is left of origin
when displacement is positive
particle is right of origin
velocity
rate of change of position with respect to time
when velocity is negative
particle is travelling left
when velocity is positive
particle is travelling right
speed=
the absolute value of velocity
acceleration
rate of change of velocity with respect to time
when acceleration is negative
acting to the left
when acceleration is positive
acting to the right
speeding up=
velocity and acceleration acting in the same direction
slowing down=
velocity and acceleration acting in opposite direction
“constant velocity”=
when a=0
“at rest”=
when v=0
“at the origin”=
when x=0
First derivative of displacement (x dot)
velocity
Second derivative of displacements (x double dot)
acceleration
second derivative shows the…
concavity
To find max/min of displacement
solve f’(t)=0 (velocity=0)
To find max/min of velocity
solve f’‘(t)=0 (acceleration=0)
