Differentiation Flashcards
Limit
the value a function approaches, when its variable tends towards a certain value
df/dx = f’(x) =
lim f(x+h) - f(x) / h
h–>0
Using first principles, find the derivative of f(x) = x^3 - 4x.
f(x+h) = (x+h)^3 - 4(x+h)
= x^3 + 3hx^2 + (3h^2 - 4)x + h^3 - 4h
lim h–> (3x^2 + 3hx + h^2 - 4)
= 3x^2 - 4
d/dx (ax^n) =
anx ^ n-1
For sums of different terms,
differentiate
each term seperately
.
.
For products or quotients,
do not differentiate
simplify first
y = x^4(x^5 - 3x^-2) + 2 differentiate
9x^8 - 6x
To find equations of tangents/normals at a point
use differentiation to find and evaluate the gradient of the curve, then substitute into y - y1 = m(x - x1)
A curve has equation y = x^3 - 5x^2 - x^3/2 + 22
tangent and normal at the point (4, -2)
tangent
y = 5x-22
normal
y = -x/5 - 6/5
A curve has equation y = a/x.
The normal to the curve at x = 3 is parallel to x - 3y + 6 = 0.
Find the value of the constant a.
m = 1/3
gradient is -3
a = 27
rate of change at f’(x) > 0
increasing
rate of change at f’(x) < 0
decreasing
rate of change at f’(x) = 0
stationary
rate of change of the gradient
second derivative
d2y/dx2
Identify where f(x) = ¼x^4 + ⅓x^3 - 3x^2 is increasing
x(x+3)(x-2)
increasing –> x > 2 , -3<x<0
d2y / dx2 < 0
local maxima
d2y / dx2 > 0
local minima
d2y / dx2 = 0
no conclusion
A point of inflection is
where the 2nd derivative changes sign - always d2y/d2x = 0
to find stationary points
find derivative
solve derivative = 0
use x to work out y
find second derivative
evaluate at each coord
For the curve y = x^3 - 12x + 3, find and classify
the coordinates of the stationary points.
d2y/dx2 = 6x
(2,-13) is a minimum
(-2,19) is a maximum
Optimisation
form equations
differentiate twice
find stationary points
classify them
answer question
A wire of length 12 cm is bent to form a rectangle.
Show that the area, A, is given by A = 6x - x2, where x is the width of the rectangle
Find the maximum possible area
area + 6x - x^2
max area when d2A/dx2 < 0
0 = 6-2x , x=3
d2A/dx2 = -2 < 0 so max at x+3
max area +6x3-9 = 9cm
An open cuboid measures x by 2x by h units.
Its inner surface area is 12 units.
Show that the volume V is
given by V = ⅔x(6 - x^2)
Find the exact value of x for
which V is maximised
SA = 2xh + 4xh + 2x^2
12 = 6xh + 2x^2
h + 12-2x^2 / 6x
vol = 2x^2h = 2/3x (6-x^2)
max V –> d2V/dx2 < 0
dV/dx = 0 = 4-2x^2
x = +- root 2
d2V/dx^2 = -4x
will be negative for x = root 2