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) 
