Calculus Differentiation (Sample)

0.26 MB·Modern Syriac

by  Ashur Cherry

ܵ ܵ ܵ ܲ
ܐܦܠܘܼ ܚܕ ܐܬ̣ ܘܼ ܝܢܡܼ
ܕܝ
(ܓ) = 0
ܕܟ
ܕ
′
[ܓܦ(ܟ)] = ܓܦ (ܟ)
ܕܟ
ܕ
′ ′
[ܦ(ܟ) + ܩ(ܟ)] = ܦ (ܟ) + ܩ (ܟ)
ܕܟ
ܕ
′ ′
[ܦ(ܟ) − ܩ(ܟ)] = ܦ (ܟ) − ܩ (ܟ)
ܕܟ
ܕ
′ ′
[ܦ(ܟ) × ܩ(ܟ)] = ܦ(ܟ) × ܩ (ܟ) + ܦ (ܟ) × ܩ(ܟ)
ܕܟ
′ ′
ܕ ܦ(ܟ) ܩ(ܟ) × ܦ (ܟ) − ܦ(ܟ) × ܩ (ܟ)
[ ] =
2
ܕܟ ܩ(ܟ) [ܩ(ܟ)]
ܕ
′ ′
[ܦ(ܩ(ܟ))] = ܦ (ܩ(ܟ)) × ܩ (ܟ)
ܕܟ
ܕ
ܢ ܢ−1
(ܟ ) = ܢܟ
ܕܟ
ܕ
(eܟ) = eܟ
ܕܟ
ܕ
ܟ ܟ
(ܐ ) = ܐ ln|ܟ|
ܕܟ
ܕ 1
ln|ܟ| =
ܕܟ ܟ
ܕ 1
(log ܟ) =
ܕܟ ܐ ܟlnܐ
ܵ ܵ ܵ ܲ
ܐܦܠܘܼ ܚܕ ܐܬ̣ ܘܼ ܝܢܡܼ
ܕ
(sin(ܟ)) = cos(ܟ)
ܕܟ
ܕ
(cos(ܟ)) = −sin(ܟ)
ܕܟ
ܕ
(tan(ܟ)) = sec2(ܟ)
ܕܟ
ܕ
(csc(ܟ)) = −csc(ܟ)cot(ܟ)
ܕܟ
ܕ
(sec(ܟ)) = sec(ܟ)tan(ܟ)
ܕܟ
ܕ
(cot(ܟ)) = −csc2(ܟ)
ܕܟ
ܕ 1
(sin−1(ܟ)) =
ܕܟ
2
√1 − ܟ
ܕ 1
(cos−1(ܟ)) = −
ܕܟ
2
√1 − ܟ
ܕ 1
(tan−1(ܟ)) =
2
ܕܟ 1 + ܟ
ܕ 1
(csc−1(ܟ)) = −
ܕܟ
2
ܟ√ܟ − 1
ܕ 1
(sec−1(ܟ)) =
ܕܟ
2
ܟ√ܟ − 1
ܵ ܵ ܵ ܲ
ܐܦܠܘܼ ܚܕ ܐܬ̣ ܘܼ ܝܢܡܼ
ܕ 1
(cot−1(ܟ)) = −
2
ܕܟ 1 + ܟ
ܕ
(sinh(ܟ)) = cosh(ܟ)
ܕܟ
ܕ
(cosh(ܟ)) = sinh(ܟ)
ܕܟ
ܕ
(tanh(ܟ)) = sech2(ܟ)
ܕܟ
ܕ
(csch(ܟ)) = −csch(ܟ)coth(ܟ)
ܕܟ
ܕ
(sech(ܟ)) = −sech(ܟ)tanh(ܟ)
ܕܟ
ܕ
(coth(ܟ)) = −csch2(ܟ)
ܕܟ
ܕ 1
(sinh−1(ܟ)) =
ܕܟ
2
√1 + ܟ
ܕ 1
(cosh−1(ܟ)) =
ܕܟ
2
√ܟ − 1
ܕ 1
(tanh−1(ܟ)) =
2
ܕܟ 1 − ܟ
ܕ 1
(csch−1(ܟ)) = −
ܕܟ
2
|ܟ|√ܟ + 1
ܵ ܵ ܵ ܲ
ܐܦܠܘܼ ܚܕ ܐܬ̣ ܘܼ ܝܢܡܼ
ܕ 1
(sech−1(ܟ)) = −
ܕܟ
2
ܟ√1 − ܟ
ܕ 1
(coth−1(ܟ)) =
2
ܕܟ 1 − ܟ