Calculus:Derivative of Trigonometric Functions

1. $f(sinx)'=cos(x)$
2. $f(cosx)'=-sin(x)$
3. $f(tanx)'=sec^{2}(x)$
4. $f(cotx)'=-csc^{2}(x)$
5. $f(cscx)'=-csc(x)cot(x)$
6. $f(secx)'=sec(x)tan(x)$

How do we get the equation

$f(sinx)'=cos(x)$ and $f(cosx)'=-sin(x)$ you only can tell by looking at the graph so we will skip it to.

tan x

So we will started with $f(tanx)'=sec^{2}(x)$

We know that $tanx={sinx \over cosx}$ If you have learn trigonometry then.

$f({sinx \over cosx})'$ by using the Quotient Rule $({f \over g})'={(gf'-fg') \over g^{2}}$

$f({sinx \over cosx})'={{cosx(sinx)'-sinx(cosx)'} \over {cosx}^{2}}$

$f({sinx \over cosx})'={{cosxcosx-sinx(-sinx)} \over {cosx}^{2}}$

$f({sinx \over cosx})'={{{cos}^{2}x+{sin}^{2}x} \over {cosx}^{2}}$

Failed to parse (syntax error): {\displaystyle {cos}^2 x + {sin}^2 x} = 1}

$f({sinx \over cosx})'={1 \over {cosx}^{2}}$