# Derivative Table

Sub Topics
The derivative table highlights the important formulaes that are necessary for solving problems that change with respect to the variable for a given application it involves rate change of a particular situation.

## Trig Derivative Table

 Trigonometric Function Derivative of the Trigonometric function 1 Sin ( θ ) cos ( θ ) 2 cos ( θ ) sin ( θ ) 3 tan ( θ ) sec2 ( θ ) 4 cot ( θ ) csc2 ( θ ) 5 sec ( θ ) Sec ( θ ) x tan ( θ ) 6 csc ( θ ) Csc( θ ) x cot( θ )

Let us see proofs of the derivatives of the trigonometric functions

Theorem: 1

The differentiation of sin ( θ ) with respect to θ is cos ( θ )
That is $\frac{d}{d\theta}$ (sin θ) = cos θ

Proof:

Let f ( θ ) =  sin ( θ ). Then f(θ + h) = sin (θ + h), we know

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [f(θ + h) – f(θ)]  h

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [sin(θ + h) – sin (θ)]  h

Using the formula of Sin C + Sin D = 2 sin $\frac{(C - D)}{2}$ * cos $\frac{(C + D)}{2}$, we get

f( θ ) = Lim h $\rightarrow$ 0  [2 sin($\frac{h}{2}$) * $\frac{cos ( 2θ + h )}{2}$]  h

f( θ ) = Lim h $\rightarrow$ 0  [2 sin($\frac{h}{2}$) * $\frac{cos (θ + h)}{2}$]  2 ($\frac{h}{2}$)

f( θ ) = Lim h $\rightarrow$ 0 cos ($\frac{θ + h}{2}$) * Lim h $\rightarrow$ 0 [sin($\frac{h}{2}$) / ($\frac{h}{2}$)]    [since, Lim x $\rightarrow$ 0 ($\frac{sinx}{x}$) = 1]

$\frac{d}{d\theta}$ f( θ ) = cos θ  * 1

$\frac{d}{d\theta}$ sin ( θ ) = cos θ

Theorem: 2

The differentiation of cos ( θ ) with respect to θ is - sin ( θ
That is $\frac{d}{d\theta}$ (cos θ) = - sin θ

Proof:

Let f ( θ ) =  cos ( θ ). Then f(θ + h) = cos (θ + h), we know

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [f(θ + h) – f(θ)]  h

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [cos (θ + h) – cos ( θ )]  h

Using the formula of Cos C - Cos D = - 2 sin$\frac{(c + D)}{2}$ * sin $\frac{(c - D)}{2}$, we get

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [- 2 sin $\frac{(2θ + h)}{2}$) * sin $\frac{(h)}{2}$]  h

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0 - sin $\frac{(θ + h)}{2}$) * Lim h $\rightarrow$ 0 [sin $\frac{(h/2)}{(h/2)}$] [since, Lim x $\rightarrow$ 0 ($\frac{sinx}{x}$) = 1]

$\frac{d}{d\theta}$ f( θ ) = - sin θ  * 1

$\frac{d}{d\theta}$ cos ( θ ) = - sin θ

Theorem: 3

The differentiation of tan ( θ ) with respect to θ is  sec2 ( θ )
That is $\frac{d}{d\theta}$ ( tan θ ) = - sec2 θ

Proof:

Let f ( θ ) =  tan ( θ ). Then f(θ + h) = tan (θ + h), we know
$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [f(θ + h) – f(θ)]  h

f( θ ) = Lim h $\rightarrow$ 0  [tan (θ + h) – tan ( θ )]  h

Writing each tan (x) term as $\frac{sin(x)}{cos(x)}$ and then simplifying the fractions by taking L.C.D we get

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [ sin(θ + h). cos θ – cos (θ + h). sin θ]  h * cos θ * cos (θ + h)

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0 [1  cos θ * cos (θ + h)] * Lim h $\rightarrow$ 0 [$\frac{sin(h)}{h}$]   [since, Lim x $\rightarrow$ 0 ($\frac{sinx}{x}$) = 1]

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0 [1   cos θ * cos (θ + h)] * 1

$\frac{d}{d\theta}$ f( θ ) = 1   cos θ * cos θ

$\frac{d}{d\theta}$ f( θ ) = 1 cos2  θ

$\frac{d}{d\theta}$ f( θ ) =  sec2  θ

$\frac{d}{d\theta}$ tan ( θ ) = sec2 θ

Theorem: 4

The differentiation of cot ( θ ) with respect to θ is  - csc2 ( θ )
That is $\frac{d}{d\theta}$   ( cot θ ) = - csc2 θ

Proof:

Let f ( θ ) =  cot ( θ ). Then f(θ + h) = cot (θ + h), we know

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [f(θ + h) – f(θ)]  h

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [cot (θ + h) – cot ( θ )]  h

Writing each cot (x) term as cos (x) / sin (x) and then simplifying the fractions by taking L.C.D we get

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [ cos (θ + h). sin θ – sin (θ + h). cos θ]  h * sin θ * sin (θ + h)

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0 [- 1  sin θ * sin (θ + h)] * Lim h $\rightarrow$ 0 [$\frac{sin(h)}{h}$]   [since, Lim x $\rightarrow$ 0 ($\frac{sinx}{x}$) = 1]

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0 [- 1   sin θ * sin (θ + h)] * 1

$\frac{d}{d\theta}$ f( θ ) = - 1   sin θ * sin θ

$\frac{d}{d\theta}$ f( θ ) = - 1   csc2  θ

$\frac{d}{d\theta}$ f( θ ) =  - csc2  θ

$\frac{d}{d\theta}$ cot ( θ ) = - csc2 θ

Theorem: 5

The differentiation of sec ( θ ) with respect to θ is  sec ( θ ) * tan ( θ )
That is $\frac{d}{d\theta}$   ( sec θ ) =  sec θ * tan ( θ )

Proof:

Let f ( θ ) =  sec ( θ ). Then f(θ + h) = sec (θ + h), we know
$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [f(θ + h) – f(θ)]  h

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [sec (θ + h) – sec ( θ )]  h

Writing each sec (x) term as $\frac{1}{cos(x)}$ and then simplifying the fractions by taking L.C.D we get

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [ cos θ – cos (θ + h)]  h * cos θ * cos (θ + h)

Using the formula of Cos C - Cos D = 2 sin $\frac{(C + D)}{2}$ * sin $\frac{(D - C)}{2}$, we get

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [2 sin($\frac{(2θ + h)}{2}$) * sin $\frac{(h)}{2}$]  h * cos θ * cos (θ + h)

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0 [sin $\frac{(2 θ + h)}{2}$  cos θ * cos (θ + h)] * Lim h $\rightarrow$ 0 [$\frac{sin\frac{h}{2}}{\frac{h}{2}}$]

[Since, Lim x $\rightarrow$ 0 (sinx / x) = 1]

$\frac{d}{d\theta}$ f( θ ) = sin θ   cos θ * cos θ

$\frac{d}{d\theta}$ f( θ ) = tan θ  cos θ

$\frac{d}{d\theta}$ f( θ ) = tan θ  * sec  θ

$\frac{d}{d\theta}$ sec ( θ ) = tan θ  * sec  θ

Theorem: 6

The differentiation of csc ( θ ) with respect to θ is  - csc ( θ ) * cot ( θ )
That is $\frac{d}{d\theta}$ ( css θ ) =  - csc ( θ ) * cot ( θ )

Proof:

Let f ( θ ) =  csc ( θ ). Then f(θ + h) = csc (θ + h), we know
$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [f(θ + h) – f(θ)]  h

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [csc (θ + h) – csc ( θ )]  h

Writing each csc (x) term as $\frac{1}{sin ( x )}$ and then simplifying the fractions by taking L.C.D we get

$\frac{d}{d\theta}$ f( θ ) = Lim h $\rightarrow$ 0  [ sin θ – sin (θ + h)]  h * sin θ * sin (θ + h)

Using the formula of Sin C + Sin D = 2 sin $\frac{(C - D)}{2}$ * cos $\frac{(C + D)}{2}$, we get

$\frac{d}{d\theta}$ f( θ ) = - Lim h $\rightarrow$ 0 [cos $\frac{(2 θ + h)}{2}$  sin θ * sin (θ + h)] * Lim h $\rightarrow$ 0 [$\frac{sin\frac{h}{2}}{\frac{h}{2}}$]
[Since, Lim x $\rightarrow$ 0 ($\frac{sinx}{x}$) = 1]

$\frac{d}{d\theta}$ f( θ ) = - cos θ  sin θ * sin θ

$\frac{d}{d\theta}$ f( θ ) = - cot θ   sin θ

$\frac{d}{d\theta}$ f( θ ) = - cot θ  * csc  θ

$\frac{d}{d\theta}$ csc ( θ ) = -  csc  θ * cot θ