Freddys70014@students.laalliance.org: Created page with " $\mathbf{Chapter 3 Section 2}$
${\frac{d}{dx}} [sin(x)] = cos(x) \qquad {\frac{d}{dx}}[csc(x)]= -csc(x) \cdot cot(x)$
${\frac{d}{dx}} [cos(x)] = -sin(x) \qquad {\frac{d}{dx}}[sec(x)]= sec(x) \cdot tan(x)$
${\frac{d}{dx}} [tan(x)] = sec^2(x)\qquad {\frac{d}{dx}}[cot(x)]= -csc^2(x)dr$
$\mathbf{\color{Blue}{Examples}}$
$\mathbf{Ex.2}$
$f(x)=\frac{sec(x)}{1+tan(x)}$
"

2023-05-02T18:08:49Z

Created page with "<math>\mathbf{Chapter 3 Section 2}</math> <math>{\frac{d}{dx}} [sin(x)] = cos(x) \qquad {\frac{d}{dx}}[csc(x)]= -csc(x) \cdot cot(x)</math> <math>{\frac{d}{dx}} [cos(x)] = -sin(x) \qquad {\frac{d}{dx}}[sec(x)]= sec(x) \cdot tan(x)</math> <math>{\frac{d}{dx}} [tan(x)] = sec^2(x)\qquad {\frac{d}{dx}}[cot(x)]= -csc^2(x)dr</math> <math>\mathbf{\color{Blue}{Examples}}</math> <math>\mathbf{Ex.2}</math> <math>f(x)=\frac{sec(x)}{1+tan(x)}</math> "

New page

<math>\mathbf{Chapter 3 Section 2}</math> 

<math>{\frac{d}{dx}} [sin(x)] = cos(x) \qquad {\frac{d}{dx}}[csc(x)]= -csc(x) \cdot cot(x)</math> 
<math>{\frac{d}{dx}} [cos(x)] = -sin(x) \qquad {\frac{d}{dx}}[sec(x)]= sec(x) \cdot tan(x)</math> 
<math>{\frac{d}{dx}} [tan(x)] = sec^2(x)\qquad {\frac{d}{dx}}[cot(x)]= -csc^2(x)dr</math> 
<math>\mathbf{\color{Blue}{Examples}}</math> 
<math>\mathbf{Ex.2}</math> 
<math>f(x)=\frac{sec(x)}{1+tan(x)}</math>

2024/G10/12 - Revision history