Unit 1 Test

Calculus

limit formula 1 (specific point)

f’(a) = lim f(x) - f(a)/ x - a, x≠a

x→a

limit formula 2 (general formula)

f’(x)=lim f(x+h) - f(x)/h, h≠0

h→0

the derivative can be calculated at…

…continuous points

the derivative does not exist at…

… asymptotes, jump discontinuity, removal discontinuity, holes

The Power Rule of Differentiation (proof)

f(x)=x^n P1 (a, a^n) P2(x, x^n)

f’(x)=lim x^n - a^n/ x - a

x→a

=lim (x - a)\[x^n-1 + xa^n-2 + x^n-3 a^2 +… + xa^n-2 + a^n-1\] x≠a

x→a

=lim \[x^n-1 + x^n-2 a + x^n-3 a^2 + x^n-4 a^3 + … + xa^n-2 + a^n-1\] sub x=a

x→a

=\[a^n-1 +a^n-2 a + a^n-3 a^2 + a^n-4 a^3 + … + aa^n-2 + a^n-1\]

= a^n-1 + a^n-1 + a^n-1 + a^n-1 + … + a^n-1 + a^n-1

= na^n-1

= nx^n-1

The Power Rule

if f(x)= x^n, f’(x) = nx^n-1

if y=x^n, dy/dx = nx^n-1 if d/dx(x^n) = nx^n-1

The Product Rule

y=f(x) g(x) y= f’(x)g(x) + g’(x)f(x)

The Product Rule (proof)

f’(x)= lim f(x+h) - f(x)/h g’(x)= lim g(x+h) - g(x)

h→0. h→0

=lim f(x+h) g(x+h) - f(x+h) g(x) + f(x+h) g(x) + f(x+h) g(x) - f(x) g(x)

h→0

= lim f(x+h)\[ g(x+h) - g(x)/h\] + lim g(x)\[ f(x+h) - f(x)/h\]

h→0 h→0

=f(x)\[g’(x)\] + g(x)\[f’(x)\]

=f’(x) g(x) + g’(x) f(x)

The Quotient Rule

f’(x) g(x) - g’(x) f(x)/\[g(x)\]^2

The Quotient Rule

if y=f(x)/g(x)

g(x) y = f(x)

g’(x) y + dy/dx g(x) = f’(x)

dy/dx = f’(x) - g’(x) (f(x)/g(x)) / g(x)

dy/dx = f’(x) (g(x)/g(x)) - g’(x) (f(x)/g(x)) / g(x)

dy/dx = f’(x)g(x) - g’(x)f(x) / \[g(x)\]^2

Limits…

can exist at continuous points & removal discontinuity - cannot exist at vertical asymptotes & jump discontinuity

If lim f(x) ≠ lim f(x) …

x→ a^- x→ a^+

it is a jump discontinuity

If f(a) ≠ lim f(x) ≠ lim f(x) …

x→ a^- x→ a^+

it is a removal discontinuity

If f(a) = lim f(x) = lim f(x) …

x→a^- x→a^+

it is continuous