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Derivatives

rename
Updated 2009-02-04 22:07

Derivatives (for each below, left side = right side)

Question Answer
f'=df/dxlim h->0 [f(x+h)-f(x)]/[(x+h)-x]
f'=df/dxlim h->0 [f(x+h)-f(x-h)]/[(x+h)-(x-h)]
(c f)'c f' (when c is a constant)
(f + g)'f' + g'
(f g)'f' g + f g'
memorize

 

Question Answer
(f g h)'f' g h + f g' h + f g h'
(1/f)'-f'/f2 (for f≠0)
(f/g)'(f' g - f g')/g2 (for g≠0)
(fg)'fg (g' ln f + g f'/f) (for f≠0)
c'0
x'1
memorize

 

Question Answer
f(g(x))'(df/dg)(dg/dx)
g(f(x))'(dg/df)(df/dx)
f(g(h(x)))'(df/dg)(dg/dh)(dh/dx)
f(h(g(x)))'(df/dh)(dh/dg)(dg/dx)
g(f(h(x)))'(dg/df)(df/dh)(dh/dx)
g(h(f(x)))'(dg/dh)(dh/df)(df/dx)
h(f(g(x)))'(dh/df)(df/dg)(dg/dx)
h(g(f(x)))'(dh/dg)(dg/df)(df/dx)
memorize

 

Question Answer
(c x)'c
|x|'x/|x|=sgn x for x≠0
(xc)'c xc-1
(1/x)'=(x-1)'-x-2=-1/x2
(1/xc)'=(x-c)'-c x-c-1=-c/xc+1
(√x)'=(x1/2)'(1/2) x-1/2=1/(2 √x) for x>0
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Question Answer
(cx)'cx ln c for c>0
(ex)'ex
(logcx)'1/(x ln c) for c>0, c≠1
(ln x)'1/x for x>0
(ln |x|)'1/x
(xx)'xx (1 + ln x)
(Γ(x))'0 tx-1 e-t ln t dt
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Question Answer
(sin x)'cos x
(cos x)'-sin x
(tan x)'sec2x = 1/cos2x
(sec x)'sec x tan x
(csc x)'-csc x cot x
(cot x)'-csc2x = -1/sin2x
memorize

 

Question Answer
(arcsin x)'1/√1 - x2
(arccos x)'-1/√1 - x2
(arctan x)'1/(1 + x2)
(arcsec x)'1/[|x| √x2 - 1]
(arccsc x)'-1/[|x| √x2 - 1]
(arccot x)'-1/(1 + x2)
memorize

 

Question Answer
(sinh x)'cosh x = (ex + e-x)/2
(cosh x)'sinh x = (ex - e-x)/2
(tanh x)'sech2x
(sech x)'-tanh x sech x
(csch x)'-coth x csch x
(coth x)'-csch2x
memorize

 

Question Answer
(arcsinh x)'1/√x2 + 1
(arccosh x)'1/√x2 - 1
(arctanh x)'1/(1 - x2)
(arcsech x)'-1/[x √1 - x2]
(arccsch x)'-1/[x √1 + x2]
(arccoth x)'1/(1 - x2)
memorize

 

Reference: http://en.wikipedia.org/wiki/Table_of_derivatives

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