Derivatives of Trigonometric functions.





As you know,

The functions SINE x(sin x) , CO-SECANT x(cos x) , TANGENT x(tan x), CO-SECANT x(csc x), SECANT x(sec x) and COTANGENT x(cot x)  are called trigonometrical functions.

You can learn more about these functions by searching about it in the search box above.

We are going to learn and prove ,what are the Derivatives of these Trigonometrical Functions , here.

a> Derivative of sin x:

The derivative of sin x is:

\dfrac{d}{dx}\sin x = \cos x

Proof:

Ley y=SIN x and let this be equation (i)

and let  \Delta x be a small increment in x and \Delta y be the corresponding small increment in y,

Then we can write:

 y + \Delta y = \sin (x+ \Delta x) and let it be equation (ii).

Now if we subtract equation (i) from equation (ii), we can get:

\Delta y = \sin (x+ \Delta x) - \sin x

Now using the trigonometrical formula ( sin c – sin d = 2.sin((c-d)/2).cos((c+d)/2)

We get:

\Delta y = 2 \times \sin \frac{\Delta x}{2} \times \cos \frac{2x+\Delta x}{2}

Now dividing both side of above equation by \Delta x we get:

\dfrac{\Delta y}{\Delta x} = \dfrac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}\times\cos\frac{2x+\Delta x}{2}

Thus:

\dfrac{d}{dx}y = \dfrac{d}{dx}\sin x = \displaystyle\lim_{{\Delta x}\to 0}\left(\dfrac{\sin \frac{\Delta x}{2}}{\frac{\Delta x}{2}}\times\cos\frac{2x+\Delta x}{2}\right)

or, \dfrac{d}{dx}\sin x = 1 \times \cos \frac{2x}{2} = \cos x

b> Derivative of cos x.

The derivative of cos x is:

\frac{d}{d x}\cos x = -\sin x

Proof:

Let y=cos x

and from trigonometric relations we also know:

\cos x = \sqrt{1-\sin ^{2}x}

So,

\frac{d }{d x} y = \frac{d }{d x}\left(\sqrt{1-\sin ^{2}x}\right)

Now using chain rule :

\frac{d }{d x} y = \frac{d }{d \left(1-\sin ^{2}x\right)} \left(1-\sin ^{2}x\right)^{\frac{1}{2}} \times \frac{d }{d x}\left(1-\sin ^{2}x\right)

or, \frac{d }{d x} y = \frac{0-2\times\sin x\times\cos x}{2\times \sqrt{1-\sin ^{2}x}} = -\sin x

Thus , \frac{d }{d x} \cos x = -\sin x

c> Derivative of  tan x.

The derivative of  tan x is:

\dfrac{d}{dx}\tan x = \sec ^{2}x

Proof:

\dfrac{d}{dx}\tan x = \dfrac{\sin x}{\cos x}

Now using the quotient rule :

or,
\dfrac{d }{dx}\tan  x =\dfrac{\cos ^2 x + \sin ^2 x}{\cos ^{2}x}

Thus,

\dfrac{d}{dx}\tan x = \sec ^{2}x

d> Derivative of csc x , secx and cot x.

Using the relation:

csc x= 1/sin x

sec x= 1/cos x

and

cot x = 1/tan x

and then using the quotient rule, we can find the:

Derivative of csc x:

\frac{d }{d x}\csc x = - \csc x \times \cot x

Derivative of sec x:

\frac{d }{d x}\sec x = \sec x \times \tan x

Derivative of cot x:

\frac{d }{d x}\cot x = -\csc ^{2} x

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