Derivative of Tan and Tan-1

Derivative of Tan:
We know that in Trignometric functions, tan x = sinx/cosx
Let us derive tan x, with respect to x
Here, f(x) = sin x and g(x) = cos x
Using  Quotient Rule for Derivatives,



 d/dx  [f(x)/g(x) ]=[g(x)  d/dx  f(x)- f(x)  d/dx  g(x)]/[g(x)]^2
=   [cos x  d/dx  sin x-sin x  d/dx  cos x]/(cos x )^2
[ d/dx  sin x=cos x    and d/dx  cos x= -sin x ]

=  [cos x .cos x- sin x.  (-sin x)  ]/(cos x )^2

=  [cos ^2  x+  sin ^2 x]/cos^2 x

                = [cos^2 x/cos^2 x +sin^2 x/cos^2 x ]                  [we have,sin^2 x/cos^2 x =tan^2 x  ]

=  1+tan^2 x
           
= sec^2 x
 
       d/dx  tan x= sec ^2 x


Derivative of Tan Inverse:
To find the derivative of an inverse of tangent function, let us take
 y = tan-1x
which would give us,
tan y = tan (tan-1x)
tan y = x            
      Differentiating on both sides with respect to y
d/dy  tan y=dx/dy
    Using the chain rule, we get
d/dy  (tan (y) )  dx/dy=1
1/cos^2 y   dy/dx=1
dy/dx=cos^2 y
y’ = cos2(tan x)                     [y=tan x]
Let us now consider a right triangle,

In the above triangle, tan (y) = x  

 y = arc tan(x)

According to the Pythagorean Theorem, hypotenuse h = v(x^2+1)

Now, we can compute:

cos y=1/v(x^2+1)

Squaring on both sides, we get

cos^2 y=[1/v(x^2+1)]^2

                             cos^2 y=1/(x^2+1)

So, dy/dx=1/(x^2+1)                                [dy/dx=d/dx  tan^(-1)  x,y=tan^(-1) x   ]  

and hence,  d/dx  tan^(-1)  x =1/(x^2+1)

Derivative of Tan x or[d/dx  tan x]:

We know that, tan x = sinx/cosx

Let us derive tan x, with respect to x

Using the Quotient Rule for Derivatives

 d/dx  [f(x)/g(x) ]=[g(x)  d/dx  f(x)- f(x)  d/dx  g(x)]/[g(x)]^2 

Here, f(x) = sin x and g(x) = cos x

=   [cos  x  d/dx  sin x-sin x  d/dx  cos x]/(cos x )^2

[ d/dx  sin  x=cos x    and d/dx  cos x= -sin x ]

 

=  [cos  x .cos x- sin x.  (-sin  x)  ]/(cos x )^2 

=  [cos ^2  x+  sin ^2 x]/cos^2 x

                      = [cos^2 x/cos^2 x +sin^2 x/cos^2 x ]                  [we have,sin^2 x/cos^2 x =tan^2 x  ]

=  1+tan^2 x

             

= sec^2 x   

       d/dx  tan x= sec ^2 x

Derivative of Tan x Inverse [tan-1(x)] or arc tan(x)):

To find the derivative of an inverse of tangent function, let us take

 y = tan-1x

tan y = tan (tan-1x)

tan y = x              

      Differentiating on both sides with respect to y

d/dy  tan y=dx/dy

    Using the chain rule, we get

d/dy  (tan (y) )  dx/dy=1

1/cos^2 y   dy/dx=1

dy/dx=cos^2 y

y’ = cos2(tan x)                     [y=tan x]

consider a right triangle,

 In the above triangle, tan (y) = xy = arc tan(x)According to the Pythagorean Theorem, h = v(x^2+1)Now, we can compute: cos y=1/v(x^2+1)From this, we get cos^2 y=[1/v(x^2+1)]^2                             cos^2 y=1/(x^2+1)     [cos^2 y=dy/dx] dy/dx=1/(x^2+1)                                [dy/dx=tan^(-1) x ]    d/dx tan^(-1)  x =1/(x^2+1)   or  d/dx arc tan(x) = 1/(x^2+1)