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
If the gradient is negative between x=a and x=b
It is a decreasing function in the interval a<x<b
Gradient of a turning point
zero
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.
