Calculus Flashcards
What does δy/δx mean?
Small change
Scalar multiple rule
d/dx[kf(x)]=kd/dx[f(x)]
Addition/subtraction rule
d/dx[f(x)+-g(x)]=d/dx[f(x)]+-d/dx[g(x)]
y=axn
dy/dx=naxn-1
What is the equation of the tangent if the value of dy/dx at the point (x,y) is m?
y-y1=m(x-x1)
What is the equation of the normal if the value of dy/dx at the point (x,y) is m?
y-y1=-(1/m)(x-x1)
Variables x and y are connected by the equation y=3x2-4x. Find, in terms of p, the approximate change in y as x increases from 5 to 5+p where p is small
dy/dx=6x-4
dy/dx(x=5)=6(5)-4=26
δy/p=26
δy=26p
y=ex
dy/dx=ex
y=ln(x)
dy/dx=1/x
y=tan(x)
dy/dx=sec2(x)
Variables x and y are connected by y=2x3+2x. Given that x increases at a rate of 0.02 units per second, find the rate of change of y when x=3
0.02dy/dx=(6x2+2)*0.02
=0.12x2+0.04
0.02dy/dx(x=3)=0.12(3)2+0.04=1.12 units per second
m(x)=f(g(x))
m|=f|(g(x))*g|(x)
m(x)=u(x)v(x)
m|(x)=u|(x)v(x)+u(x)v|(x)
m(x)=u(x)/v(x)
m|(x)=(u|(x)v(x)-u(x)v|(x))/v2(x)
At the maximum point
- dy/dx=0
- The gradient is positive to the left of the maximum and negative to the right
- d2y/dx2<0
At the minimum point
- dy/dx=0
- The gradient is negative to the left of the minimum and positive to the right
- d2y/dx2>0
dy/dx=xn
y=xn+1/n+1 + c where c is a constant and n!=-1
∫kf(x)dx
k∫f(x)dx where k is a constant
∫[f(x)+-g(x)]dx
∫f(x)dx+-∫g(x)dx
∫(ax+b)ndx
(ax+b)n+1/a(n+1) + c where n!=-1 and a!=0
∫1/(ax+b)dx
(1/a)ln|ax+b| + c
∫eax+bdx
(1/a)eax+b + c
∫sec2(x)dx
tan(x) + c
Find ∫21((x5+3)/x2)dx
=∫21x3+3x-2 dx
= [x4/4+3x-1/-1]21
= [x4/4-3/x]21
= 24/4-3/2-(14/4-3/1)=21/4