Differentiation Flashcards
δχ and δy
small changes in x and y between 2 points
dχ and dy
basically δχ and δy but more accurate in terms of the gradient and so more helpful
y = kx^n
(dy)/(dχ) = knχ^(n-1)
Ok, it’s unclear but this legitimately might only work on a term by term basis.
This would kind of make sense as each term is adding upon the gradient to a specific level and so this conversion on a term by term basis would be proportional to the amount that they are adding to the gradient.
How to actually differentiate but with an example to show it
y = kx^n
(dy)/(dχ) = knχ^(n-1)
y=2x^2+5x+7
(dy)/(dχ) = 22x^(2-1)+51x^(1-1) +7*0x^-1
=4x+5+0
=4x+5
How do you find out the gradient of a curve at a coordinate?
You differentiate the entire function to make a new function.
You plug in the x coordinate.
Congrats.
What’s the normal?
It’s the perpendicular line of the tangent that goes through the same spot on the curve.
If the gradient is positive between x=a and x=b
It is an increasing function in the interval a<x<b
Gradient of a turning point
zero
If the gradient is negative between x=a and x=b
It is a decreasing function in the interval a<x<b
Maximum point
Increasing function to the left decreasing function to the right
Minimum point
Decreasing function to the left Increasing function to the right
Gradient Function graph
Differentiate the dy/dx and plot it.
Alternatively when turning point, it touches the x axis. When steepest between turning points give it a turning point.
Increasing- positive y
decreasing- negative y
Differentiate dy/dx
d2y/dx^2
dy/dx = f’(x) so…
d2y/dx^2= f’‘(x)
Plug x coordinate of turning point into d2y/dx^2
if negative- max
if positive- min
if zero- carefully take measurements of gradient from either side of it while not crossing over another turning point and figure it out from there.
