dy/dx = lim (h -> 0) [(f(x + h)- f(x)]/h formula for derivative as a limit d/dx [sinh x] = d/dx [csc x] = d/dx [arccot x] =

formula for derivative as a limit

calculus 1&2 review (Limits, drivatives, integration, etc.)

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Calculus - Geometry

user_lira_cofre Created by 6 mon ago

Cards in this deck(46)

it describes a function that approches as its input and it also approaches as ts input and it also approaches a certain value or infinity

lim (x->a) [f(x)/g(x)] = lim (x->a) f'(x)/g'(x)

dy/dx = lim (h -> 0) [(f(x + h)- f(x)]/h

the instantaneous rate of change of a function in relation to another variable.

any number or any variable that doesn't respect to it will remain "0" e.g. d/dx (5) = 0 or d/dx (c) = 0

d/dx[cf(x)]=cf'(x)

d/dx[f(x)+g(x)]=f'(x)+g'(x)

d/dx [f(x) - g(x)] = f'(x) - g'(x)

d/dx [f(x)g(x)] = f'(x)g(x)+f(x)g'(x)

d/dx [f(x)/ g(x)] = (g(x) f'(x) - f(x) g'(x))/ [g(x)]^2

d/dx f[g(x)] = f'(g(x)) g'(x)

cos x

-sin x

sec^2 x

sec x tan x

-csc^2 x

-csc x cot x

cosh x

sinh x

sech^2 x

-sech x tanh x

-csch x coth x

-csch^2 x

1/√(1-x^2)

-1/√(1-x^2)

1/(1 + x^2)

1/|x|√x^2-1

1/(1 + x^2)

1/(x ln a)

1/x

b^x ln b

ae^ax

-1/x^2

1/2√x

d^2 y/dx^2 = d/dx (dy/dx)

how quickly something changes over time

f(b)-f(a)/b-a

the expression of finding the maximum or minimum value of a function and often a subject to constraints

L(x) = f(a) + f'(a)(x - a)

a numerical value representing the area under a function's graph over a specific interval

a way to approximate the area under the curve by dividing it into simple shapes like rectangle or trapezoids and adding up their areas

L_n = ∑ f(x_i) Δx Δx = (b - a/n)

R_n = ∑ f(x_i) Δx Δx = (b - a/n)

M_n = ∑ [f(x_i-1 + x_i/2)] Δx Δx = (b - a/n)

T_n = Δx/2 ∑ [f(x_i-1 + f(x_i)] Δx = (b - a/n)

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