Math Test 2 Flashcards
Chain Rule
d/dxf(g(x)) = f’(g(x))g’(x) = df/dgdg/dx
Differentiate f(x)=sin(3x)
cos(3x)(3)
Differentiate f(x)=(x^3-2x+1)^6
6(x3-2x+1)5(3x2-2)
Implicit Differentiation
- Taking derivative of both sides using Chain rule
- Solve for dy/dx
Differentiate x^2+y^2=1, then find x where y’=1 using Algebra & Diagram
dy/dx=-2x/2y
x=+/- 1/sqrt(2)
Logarithmic Differentiation
Applying the natural logarithm to both sides of equation to make it easier to differentiate, especially in f(x)^g(x), combining Chain Rule+Implicit Differentiation+Logarithmic Rules
- Take log of both sides, simplifying using log rules
- Take d/dx of both sides (implicit differentiation+chain rule)
- Isolate for y’(x)
Differentiate x^a using logarithmic differentiation
f(x)= x^a
logf(x) = logx^a
1/f(x) * f’(x) = alogx
1/(x^a) * f’(x)=a/x
f’(x) = a/x(x^a)
f’(x)= ax^(a-1)
How to solve a related rate question?
- Draw & label a picture + Write down what is known
- Write down what you wish to find
- Write down an equation that relates the thing whose rate y (reduce to 1 variable)
- Differentiate & Solve
- Check if answer makes sense
Rate of Change of Area = L’W + LW’
Rate of Change of Focal Length of Lense
Rate of Change of radius of expanding circle
Rate of change of water level entering inverted cone
Rate of Change of angle of a moving person across a room
Wall = 5.9m
Rate of Change of Walking = 1/4m/s
x = opposite/walking distance
tangent = angle
At tangent=pi/4, what is the rate of change of tangent?
Draw label diagram
Find dtangent/dt or rate of change of tangent or f’(pi/4) f(x)=tangent
t * tan(tangent)=tx/L –> differentiate ^ solve
(1215.6)1.4
An inverted cylindrical cone, 28 m deep and 14 m in diameter across the top, is being filled with water at a constant rate of 13 m3/min. At what rate is the water rising in the tank when the depth of the water is 1m, 10m, 27m
208/pi(h)^2
Draw sin(x)=arcsin(y) & sin(y)=arcsin(x) at range[-pi/2,pi/2]<->domain[-1,1]
See Diagram
Draw cos(x)=arccos(y) & sin(y)=arccos(x) at range[-0,pi]<->domain[-1,1]
See diagram
Differentiate arcsin(x) in terms of x
d/dxsin(y)=dx/dx
cos(y)dy/dx=1
Triangle + SOH CAH TOA
dydx=11-x2
Differentiate arccos(x) in terms of x
d/dxcos(y)dy/dx=dx/dx
dy/dx=-1/sin(y)
Triangle + SOH CAH TOA
dy/dx=-1/sqrt(1-x2)
Differentiate arccos(e^2x) in terms of x
2e^(2x)/sqrt(1-e^4x)
Draw tan(x)=arctan(y) & tan(y)=arctan(x) at range[-pi/2,pi/2]<->domain[-infinity,infinity]
See diagram
Differentiate arctan(x) in terms of x
d/dxtan(y)=dx/dx
sec^2(y)dy/dx=1
dy/dx=1/sec2(y)
dy/dx=cos^2(y)
Triangle + SOH CAH TOA
dy/dx=1/1+x2
How to determine concavity?
If f(x)’‘>0 = slope is increasingly increasing = concave up = f(x) lies above its tangent lines on that interval
Diagram
If f(x)’‘<0 = slope is increasingly decreasing = concave down = f(x) lies below its tangent lines on that interval
Diagram
Degree 0 & 1(linear) approximation around x=0? Diagram?
T0(x) be the degree 0 approximation of f(x) around x=0: T0(x)=f(0)=1
Let T1(x)be the linear approximation to f(x) about x=0: T1(x)=f(0)+f’(0)(x-a)=1+1(x-0) = 1+x
The higher the degree of approximation, the better the approximation
Degree n Approximation to f(x) at x=a is ?
Tn(x)=f(a)+f’(a)(x-a)+f’‘(a)2(x-a)2+f’’‘(a)3!(x-a)3+…+f(n)(a)n!(x-a)n
Taylor vs Maclaurin Polynomial
Taylor x=a
Maclaurin x=a=0
Proof of Degree nth Approximation
Find c0,c1,c2,c3,…,cn, such that the kth derivatives match: T(k)n(x)=f(k)(a) for k=0,1,2,3,…,n
Proof of Degree 3 e^x Approximation
Let T2(x)=c0+c1(x-0)+c2(x-0)2=c0+c1x+c2x2 be the degree 2 approximation to f(x)=ex about x=0
Find c0, c1 and c2 such that
T2(0)=f(0)=1 → c0+c1(0)+c2(0)2=c0=1
T’2(0)=f’(0)=1 → c1+2c2(0)=c1=1
T’‘2(0)=f’‘(0)=1 → 2c2=1, c2=1/2
Therefore: T2(x)=c0+c1x+c2x2 = 1+1x+1/2x
cos(x) nth Degree Maclaurin Polynomial + Show Work
Tn(x)=1-x2/2!+x4/4!+…
