zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

tangent y=sqrt(2x+1)

解答

切线

y = 2 x + 1

解答

y = \frac{1}{2 a _{0} + 1} x + 2 a_{0} + 1 - \frac{a _{0}}{2 a _{0} + 1}

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, 2 a_{0} + 1)

计算

y = 2 x + 1

的斜率:

\frac{d y}{d x} = \frac{1}{2 x + 1}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{2 a _{0} + 1}

找到斜率 m=

\frac{1}{2 a _{0} + 1}

且穿过

(a_{0}, 2 a_{0} + 1)

的直线:

y = \frac{1}{2 a _{0} + 1} x + 2 a_{0} + 1 - \frac{a _{0}}{2 a _{0} + 1}

y = \frac{1}{2 a _{0} + 1} x + 2 a_{0} + 1 - \frac{a _{0}}{2 a _{0} + 1}

作图

流行的例子

inverselaplace 1/((s^2+1)(s^2+25))

in v erse l a pl a ce \frac{1}{( s ^{2} + 1 ) ( s ^{2} + 25 )}

y^'=((5x^3+y^3))/(xy^2)

y^{^{'}} = \frac{( 5 x ^{3} + y ^{3} )}{x y ^{2}}

derivative f(x)=sqrt(6x+1)

d er i v a t i v e f (x) = 6 x + 1

derivative r(x)=sin(x^2)+(1-x)/(1+x)

d er i v a t i v e r (x) = sin (x^{2}) + \frac{1 - x}{1 + x}

derivative 3arctan(x-sqrt(1+x^2))

d er i v a t i v e 3 arctan (x - 1 + x^{2})