Derivatives
Evaluate f'(1) when:
f(x) = 6x5 - x2 + 8lnx + 1
Answer: f'(1) = 36
f'(x) = 30x4 - 2x + 8/x --> f'(1) = 30 - 2 + 8 = 36
Evaluate: lim f(x) -> pi/6
f(x) = cos(x + pi/6)
Answer: 1/2
cos( pi/6 + pi/6 ) = cos(2*pi/6) = cos(pi/3) = 1/2
What are the formulas for finding the tangent line and normal line?
Tangent: y = f'(c)(x - c) + f(c)
Normal: y = (-1/f'(c))(x - c) + f(c)
Find the absolute max value for:
y = 3sinx on [0 , pi]
Answer: Abs. Max = 3
Endpoints: (0,0) and (pi,0)
y' = 3cosx --> 0 = 3cosx --> xc = pi/2 --> (pi/2 , 3)
Evaluate dy/dx at (2,2):
x2 + y2 = 8
Answer: -1
dy/dx = -x/y --> -2/2 = -1
Find dy/dx of 2x + ln(5x) - tan(x2)
Remember that d/dx(ax) = lna*ax
ln2*2x + 1/x - 2xsec2(x2)
Evaluate limit as x approaches 9:
(x2 + 2x - 35)/(x2 - 2x - 63)
Answer: DNE
[(x - 5)(x + 7)]/[(x + 7)(x - 9)] = (x - 5)/(x - 9)
(9 - 5)/(9 - 9) = 4/0 = DNE
Find the tangent line that goes through (1,4) on the function f(x) = 3x2 + x.
Answer: y = 7x - 3
c = 1 ; f(c) = 4 ; f'(c) = 7
Find the absolute min value for:
y = 2 - sinx on [0 , 3pi/2]
Answer: Abs. Min = 1
Endpoints: (0,2) and (3pi/2,3)
y' = -cosx --> 0 = -cosx --> xc = pi/2 --> (pi/2 , 1)
Evaluate (f-1)'(42) given the function:
f(x) = x5 + 10
Answer: 1/80
Inverse: x = y5 + 10 --> dy/dx = 1/(5y4)
y = 2 --> dy/dx = 1/(5(2)4) = 1/(5*16) = 1/80
Evaluate f'(0) given the function:
f(x) = (8x + 2)3
Answer: 96
f'(x) = 3(8x + 2)2(8) --> f'(x) = 24(8x + 2)2
f'(0) = 24(2)2 --> f'(0) = 24 x 4 = 96
Identify the discontinuity for:
f(x) = (3x)/(x2 - 10x + 25)
One Essential Discontinuity (asymptote)
Find the slope of the normal line at x = 2 of the function f(x) = x2 - x.
Answer: -1/3
f'(x) = 2x - 1 -> f'(2) = 3 = f'(c) -> -1/f'(c) = -1/3
True or False:
If f'(x) > 0 for all real numbers, then f(x) is a one-to-one function.
Answer: True
If the derivative is always negative, then the function is always decreasing and never increases. It passes the vertical and horizontal line test.
Find dy/dx:
2x2 - 4y2 = y
Answer: dy/dx = (4x)/(1 + 8y)
4x - 8y*(dy/dx) = 1*(dy/dx)
4x = (dy/dx)(1 + 8y) --> (4x)/(1 + 8y)
Find dy/dx: y = 5x2*sin3(x)
dy/dx = 10xsin3(x) + 15x2sin2(x)cos(x)
Evaluate: lim h -> 0 [ (root(64 + h) - 8) / h ]
Answer: 1/16
Conjugate: root(64 + h) + 8
lim h -> 0 (1)/(root(64+8) + 8) = (1)/(8+8) = 1/16
Find the tangent line that goes through (-2,-4) on the function f(x) = (2/x) - x2.
Answer: y = (7/2)x + 3
f'(x) = -2/x2 - 2x --> f'(-2) = (-1/2) + 4 = 7/2
y = (7/2)(x + 2) - 4 --> y = (7/2)x + 3
Find the critical number(xc) for the function:
y = 3x - 17lnx
Answer: xc = 17/3
y' = 3 - 17/x --> 0 = 3 - 17/x --> x = 17/3
Evaluate (f-1)'(10) given the function:
f(x) = x3 + x
Answer: 1/13
Inverse: x = y3 + y and y = 2
1 = dy/dx(3y2 + 1) --> dy/dx = (1)/(3y2 + 1)
Find dy/dx of y= cos3(1 - 2x2)
dy/dx = 12xcos2(1 - 2x2)sin(1 - 2x2)
At what x-values is the function not differentiable?
f(x) = |x - 2|/(x2 - 1)
Asymptotes: x = -1 ; x = 1
Corner: x = 2
Find the slope of the normal line at x = 3 of the function f(x) =(x2 - 1)/(x2 + 1).
Answer: -25/3
f'(x) = (4x)/((x2 + 1)2) (quotient rule)
c = 3 --> f'(c) = 3/25 --> -1/f'(c) = -25/3
Evaluate f''(pi/6) when given the function:
f(x) = sin2(x)
Answer: 1
f'(x) = 2sinxcosx (now need to do product rule)
f''(x) = 2cos2x - 2sin2x --> f''(pi/6) = 3/2 - 1/2 = 1
Find dy/dx at (3,1)
y2 + xy = 4
Answer: -1/5
2y*(dy/dx) + y + x*(dy/dx) = 0
dy/dx = (-y)/(2y + x) --> dy/dx(3,1) = -1/5