May 15, 2024
lim_(x→3) [(x^2 - 9)/(x - 3)]
:(x^2 - 9) = (x + 3)(x - 3)
.(x - 3)
, resulting in lim_(x→3) [(x + 3)]
.3 + 5 / 3 + 3 = 8 / 6 = 4 / 3
.d/dx [x^n] = nx^(n-1)
.x^4
: Result is 4x^3
.x^(-1)
: Result is -x^(-2)
reversed for reciprocal.6x^5 - 3/x^2 + 1/(2√x)
.f(x)
to be continuous at x=c
, both pieces of the function must be equal at x=c
.2cx - 6 = x^2 + cx
at x=3
.2c(3) - 6 = 9 + 3c
-> 6c - 6 = 9 + 3c
-> 3c = 15
-> c = 5
.e^(u)
derivative is e^(u) * u'
.ln(u)
derivative is u'/u
.(f*g)' = f'*g + f*g'
.e^(4x) * ln(2x+5)
:4e^(4x)(ln(2x+5)) + e^(4x)*(2/(2x + 5))
.(x^n dx) = (x^(n+1))/(n+1)
+ C.(4x^5 + x^4 - 3x^2)/x^2
(4x^3 + x^2 - 3x)
.x^4/4 + (x^3/3) - (3x^2/2) + C
.y - y1 = m(x - x1)
, where m
is slope.2x^3 + 4xy^2 + y^3 = 107
at (2,3).16x + 25y = 107
.f'(x) = lim_(h→0) [f(x+h) - f(x)] / h
.cos x
.2x√(3x^2 + 5) dx
.u = 3x^2 + 5
, then, du = 6x dx
or dx = du/(6x)
.2/3(u^(3/2)) + C
.V=πr^2h
).dh/dt = 3 ft/min
, r=3ft
, find dV/dt
.dV/dt = π * 9 * 3 = 27π ft³/min
.-x²+16x+5
has max point found by solving f'(x)=0
.1/(b-a) ∫_[a to b] f(x) dx
.f(x) = x^3 + 8x - 4
from [1, 5].59
for average value.d/dx f(g(x)) = f'(g(x)) * g'(x)
.(2x^3 - 7x^2)^8
.
8(2x^3 - 7x^2)^7 * (6x^2 - 14x)
simplified.limit
involving fractions simplifies to cancellation method.