derivative

If this was meant to be \(\frac{d\tan{2\pi x}}{dx}\)then the answer is \(2\pi(\sec{2\pi x})^2\)

Possible derivation:
d/dx(tan(2) pi x)
Factor out constants:
  =  pi (d/dx(x)) tan(2)
The derivative of x is 1:
  =  pi tan(2) 1
Simplify the expression:
Answer: | =  pi tan(2)

Possible derivation:
d/dx(tan(2 pi x))
Rewrite the expression: tan(2 pi x) = (sin(2 pi x))/(cos(2 pi x)):
  =  d/dx((sin(2 pi x))/(cos(2 pi x)))
Use the quotient rule, d/dx(u/v) = (v ( du)/( dx)-u ( dv)/( dx))/v^2, where u = sin(2 pi x) and v = cos(2 pi x):
  =  ((d/dx(sin(2 pi x))) cos(2 pi x)-(d/dx(cos(2 pi x))) sin(2 pi x))/(cos^2(2 pi x))
Using the chain rule, d/dx(sin(2 pi x)) = ( dsin(u))/( du) ( du)/( dx), where u = 2 pi x and ( d)/( du)(sin(u)) = cos(u):
  =  (-((d/dx(cos(2 pi x))) sin(2 pi x))+(d/dx(2 pi x)) cos(2 pi x) cos(2 pi x))/(cos^2(2 pi x))
Simplify the expression:
  =  ((d/dx(2 pi x)) cos^2(2 pi x)-(d/dx(cos(2 pi x))) sin(2 pi x))/(cos^2(2 pi x))
Factor out constants:
  =  (-((d/dx(cos(2 pi x))) sin(2 pi x))+2 pi (d/dx(x)) cos^2(2 pi x))/(cos^2(2 pi x))
The derivative of x is 1:
  =  (-((d/dx(cos(2 pi x))) sin(2 pi x))+1 2 pi cos^2(2 pi x))/(cos^2(2 pi x))
Using the chain rule, d/dx(cos(2 pi x)) = ( dcos(u))/( du) ( du)/( dx), where u = 2 pi x and ( d)/( du)(cos(u)) = -sin(u):
  =  (2 pi cos^2(2 pi x)--(d/dx(2 pi x)) sin(2 pi x) sin(2 pi x))/(cos^2(2 pi x))
Simplify the expression:
  =  (2 pi cos^2(2 pi x)+(d/dx(2 pi x)) sin^2(2 pi x))/(cos^2(2 pi x))
Factor out constants:
  =  (2 pi cos^2(2 pi x)+2 pi (d/dx(x)) sin^2(2 pi x))/(cos^2(2 pi x))
The derivative of x is 1:
  =  (2 pi cos^2(2 pi x)+1 2 pi sin^2(2 pi x))/(cos^2(2 pi x))
Use the Pythagorean identity: 2 pi cos^2(2 pi x)+2 pi sin^2(2 pi x) = (2 pi) (sin^2(2 pi x)+cos^2(2 pi x)) = (2 pi) (1) = 2 pi:
  =  2 pi/(cos^2(2 pi x))
Simplify the expression:
Answer: | =  2 pi sec^2(2 pi x)