泰勒 tan(x),3

解答

泰勒

tan (x), 3

tan (3) + sec^{2} (3) (x - 3) + sec^{2} (3) tan (3) (x - 3)^{2} + \frac{- 4 sec ^{2} ( 3 ) + 6 sec ^{4} ( 3 )}{6} (x - 3)^{3} + \frac{- 8 sec ^{2} ( 3 ) tan ( 3 ) + 24 sec ^{4} ( 3 ) tan ( 3 )}{24} (x - 3)^{4} + \dots

求解步骤

应用泰勒公式

= tan (3) + \frac{\frac{d}{d x} ( tan ( x ) ) ( 3 )}{1 !} (x - 3) + \frac{\frac{d ^{2}}{d x ^{2}} ( tan ( x ) ) ( 3 )}{2 !} (x - 3)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( tan ( x ) ) ( 3 )}{3 !} (x - 3)^{3} + \dots

求导数

= tan (3) + \frac{sec ^{2} ( 3 )}{1 !} (x - 3) + \frac{2 sec ^{2} ( 3 ) tan ( 3 )}{2 !} (x - 3)^{2} + \frac{- 4 sec ^{2} ( 3 ) + 6 sec ^{4} ( 3 )}{3 !} (x - 3)^{3} + \frac{- 8 sec ^{2} ( 3 ) tan ( 3 ) + 24 sec ^{4} ( 3 ) tan ( 3 )}{4 !} (x - 3)^{4} + \dots

整理后得

= tan (3) + sec^{2} (3) (x - 3) + sec^{2} (3) tan (3) (x - 3)^{2} + \frac{- 4 sec ^{2} ( 3 ) + 6 sec ^{4} ( 3 )}{6} (x - 3)^{3} + \frac{- 8 sec ^{2} ( 3 ) tan ( 3 ) + 24 sec ^{4} ( 3 ) tan ( 3 )}{24} (x - 3)^{4} + \dots

\frac{\partial}{\partial x} (8 x y^{2} z^{3})

in v erse l a pl a ce \frac{4}{2 ( s - 1 ) ^{2} + 50}

\frac{d}{d x} (12 sec (x))

\int \frac{sin ( x )}{cos ( x ) + cos ^{2} ( x )} d x

a re a 4 x^{2} - 4, x^{8} - 1