微分公式(6) 三角函數的微分
Author: 藍邦偉　老師

反三角函數的微分

1

y=sin^{-1}x

2

y=cos^{-1}x

3

4

5

6

y=tan^{-1}x

y=cot^{-1}x

y=sec^{-1}x

y=csc^{-1}x

y=sin^{-1}x

⇒siny=x

⇒cosy\cdot \frac{dy}{dx}=1

⇒\frac{dy}{dx}=\frac{1}{cosy}

$ =\sqrt{1-x²}$

y=cos^{-1}x

⇒cosy=x

⇒-siny\cdot \frac{dy}{dx}=1

⇒\frac{dy}{dx}=-\frac{1}{siny}

=-\frac{1}{\sqrt{1-x²}}

y=tan^{-1}x

⇒tany=x

⇒sec²y\cdot \frac{dy}{dx}=1

⇒\frac{dy}{dx}=\frac{1}{sec²x}

y=cot^{-1}x

⇒coty=x

=-\frac{1}{\sqrt{x²+1}}

⇒-csc²y\cdot \frac{dy}{dx}=1

⇒\frac{dy}{dx}=-\frac{1}{csc²x}

y=sec^{-1}x

⇒secy=x

⇒\frac{dy}{dx}=\frac{1}{tany\cdot secy}

⇒tany\cdot secy\cdot \frac{dy}{dx}=1

y=csc^{-1}x

⇒cscy=x

⇒\frac{dy}{dx}=-\frac{1}{coty\cdot cscy}

⇒-coty\cdot cscy \cdot \frac{dy}{dx}=1

y

1

x

$ \sqrt{1-x²} $

1

y

$ x $

$ \sqrt{1-x²} $

$ y $

1

x

$ \sqrt{1+x²} $

=\frac{1}{1+x²}

y

1

x

$ \sqrt{x²+1} $

y

x

1

$ \sqrt{x²-1} $

=\frac{1}{\frac{\sqrt{x²-1}}{x}\cdot \frac{x}{1}}

$ =\frac{1}{\sqrt{x²-1}} $

y

x

1

$ \sqrt{x²-1} $

=-\frac{1}{\sqrt{x²-1}\cdot x }

a = true
b = false
c = false
d = false
e = false
f = false
STEP = 0 (slider from 0 to 5, step 1)
Step = 0 (slider from 0 to 5, step 1)
step = 0 (slider from 0 to 5, step 1)
S = 0 (slider from 0 to 5, step 1)
s = 0 (slider from 0 to 6, step 1)
T = 0 (slider from 0 to 5, step 1)
A = (4.86, -0.4)
B: A + (3, 4)
C: A + (3, 0)
g = Segment(A, C)
h = Segment(C, B)
i = Segment(B, A)
D = (-1.48, 5.16)