Calculus:Derivative of Trigonometric Functions

equations you need to remember

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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 $({f \over g})'={(gf'-fg') \over g^{2}}$

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

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

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

${cos}^{2}x+{sin}^{2}x=1$

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

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

need to know that

${1 \over cosx}=secx$

$f(tanx)'=f({sinx \over cosx})'={1 \over {cos}^{2}x}={sec}^{2}x$

cot x

why did $f(cotx)'=-csc^{2}(x)$

first start with

$cotx={cosx \over sinx}$

so

$f(cotx)'=f({cosx \over sinx})'$

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

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

$f(sinx)'=cos(x)$ $f(cosx)'=-sin(x)$

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

$f({cosx \over sinx})'={-{sin}^{2}x-{cos}^{2}x \over {sin}^{2}x}$ but $-{sin}^{2}x-{cos}^{2}x=-1$ so $f({cosx \over sinx})'={-1 \over {sin}^{2}x}$

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

${1 \over sinx}={csc}x$

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

csc x

why did $f(cscx)'=-csc(x)cot$

In trigonometry $cscx={1 \over sinx}$

so

$f(cscx)'=f({1 \over sinx})'$

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

g = sin x

f = 1

$f({1 \over sinx})'={{sinx(1)'-1(sinx)'} \over {sin}^{2}x}$

need to know (1)'= 0

and (sin x)' = cos x

so

$f({1 \over sinx})'={{(0sinx)-cosx} \over {sin}^{2}x}$

next

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

${-cosx \over {sin}^{2}x}=-{cosx \over sinx}{1 \over sinx}$

In trigonometry

${cosx \over sinx}=cotx$ ${1 \over sinx}=cscx$

so

$f({1 \over sinx})'=-cotx*cscx$

Sec x

why did $f(secx)'=sec(x)tan(x)$

In trigonometry

$secx=(1 \over cosx)$

so we can say that

$f(secx)'=f({1 \over cosx})'$

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

g = cos x

f = 1

$({1 \over cosx})'={cosx(1)'-(1)(cosx)' \over {cos}^{2}x}$

$({1 \over cosx})'={cosx(1)'-(1)(cosx)' \over {cos}^{2}x}$

so

${cosx(1)'-(1)(cosx)' \over {cos}^{2}x}={sinx \over {cos}^{2}x}$

${sinx \over {cos}^{2}x}={{sinx \over cosx}{1 \over cosx}}$

In trigonometry

${sinx \over cosx}=tanx$

${1 \over cosx}=secx$ <\noinclude>