AP Calculus BC / Unit 2: Differentiation: Definition and Fundamental Properties
2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions
Notes
- $\displaystyle \frac{d}{dx}\tan x=\sec^2x$
- $\displaystyle \frac{d}{dx}\cot x=-\csc^2x$
- $\displaystyle \frac{d}{dx}\sec x=\sec x\tan x$
- $\displaystyle \frac{d}{dx}\csc x=-\csc x\cot x$
Practice
$\displaystyle \frac{d}{dx}\tan x$ at $x=0$ equals:
$\sec^2 0=1$.
$\displaystyle \frac{d}{dx}\sec x=$
Standard result from the quotient rule on $\displaystyle \frac{1}{\cos x}$.
Which derivative has a negative sign?
$\displaystyle \frac{d}{dx}\csc x=-\csc x\cot x$.
Ready for more? Take the Unit 2 test.