RATES OF CHANGE & THE DEFINITION
Write the limit definition of f'(x) using h.
f'(x) = lim_(h->0) [f(x+h) - f(x)] / h
If f is differentiable at x = a, must f be continuous at x = a?
Yes. Differentiable implies continuous.
Find d/dx (x^5).
5x^4
State the product rule for d/dx [u(x) -3v(x)].
u'(x) (-3)v(x) + u(x) (-3)v'(x)
Find d/dx (cos x).
-sin x
What does f'(a) represent graphically?
The slope of the tangent line to y = f(x) at x = a (also the instantaneous rate of change of f at x = a).
True or false: if f is continuous at x = a, then f is differentiable at x = a. Give an example.
False. Example: f(x) = |x| is continuous at x = 0 but not differentiable there (corner).
Find d/dx (3x^4 - 2x + 7).
12x^3 - 2
State the quotient rule for d/dx [h(x) / 3k(x)].
[h'(x) 3k(x) - h(x) 3k'(x)] / [k(x)]^2
Find d/dx (e^x + ln x).
e^x + 1/x
Find the average rate of change of f(x) = x^2 on [1, 3].
[f(3) - f(1)] / (3 - 1) = (9 - 1)/2 = 4
Name three ways a function can fail to be differentiable at a point.
Corner, cusp, vertical tangent line, or a discontinuity.
Find d/dx (4 / x^3).
Rewrite as 4x^(-3).
Derivative: -12x^(-4) = -12 / x^4
Find d/dx [x^2 (3x + 1)].
2x(3x + 1) + x^2(3) = 9x^2 + 2x
Give the derivatives of tan x, cot x, sec x, and csc x.
d/dx(tan x) = sec^2 x
d/dx(cot x) = -csc^2 x
d/dx(sec x) = sec x tan x
d/dx(csc x) = -csc x cot x
Write the alternate form of the definition of f'(a) (using x -> a).
f'(a) = lim_(x->a) [f(x) - f(a)] / (x - a)
Is f(x) = | x - 4 | differentiable everywhere?
No. f is not differentiable at x=4
Find d/dx (5 sqrt(x) - 6 / x^2).
Rewrite as 5x^(1/2) - 6x^(-2). Derivative: (5/2)x^(-1/2) + 12x^(-3) = 5 / (2 sqrt(x)) + 12 / x^3
Find d/dx [(x + 1) / (x - 1)].
[(1)(x - 1) - (x + 1)(1)] / (x - 1)^2 = -2 / (x - 1)^2
Find d/dx (x^2 e^x).
2x e^x + x^2 e^x = x e^x (x + 2)
The limit lim_(h->0) [(2+h)^3 - 8] / h is the derivative of what function at what x-value? What is its value?
f(x) = x^3 at x = 2. f'(x) = 3x^2, so f'(2) = 12.
Let f(x) = x^2 for x <= 1
and f(x) = 2x - 1 for x > 1.
Is f differentiable at x = 1? If so, find f'(1).
Continuous at 1: left = 1, right = 2(1) - 1 = 1.
Left derivative: 2x = 2.
Right derivative: 2.
They match, so f is differentiable at x = 1 and f'(1) = 2.
Find the equation of the tangent line to f(x) = x^3 - 2x at x = 2.
f(2) = 4. f'(x) = 3x^2 - 2,
so f'(2) = 10.
y - 4 = 10(x - 2), or y = 10x - 16
Find all x-values where f(x) = x / (x^2 + 1) has a horizontal tangent.
f'(x) = [(1)(x^2 + 1) - x(2x)] / (x^2 + 1)^2 = (1 - x^2) / (x^2 + 1)^2
f'(x) = 0 when 1 - x^2 = 0, so x = 1 and x = -1
Find d/dx [sin x / (1 + cos x)] and simplify.
[cos x (1 + cos x) - sin x (-sin x)] / (1 + cos x)^2
= [cos x + cos^2 x + sin^2 x] / (1 + cos x)^2
= (cos x + 1) / (1 + cos x)^2
= 1 / (1 + cos x)