How do you find the derivative of #y=tan(2x)#?

Essentially we use the rule:

#d/dx(a* tan(k*x+h))=(a*k)/(cos^2(k*x+h))#

or

#a*k*sec^2(k*x+h)#

I'll go through the steps behind it as well:

Derivitave of #y=tan(2x)#

The first step is to separate it into #sin(x)# and #cos(x)#:

#y=sin(2x)/cos(2x)#

Then we integrate using the quotient rule:

#f(x)=g(x)/(h(x)),# # f'(x)= (g'(x)*h(x)-h'(x)*g(x))/(h(x))^2#

#y'=(cos(2x)xx2cos(2x) - sin(2x)xx-2sin(2x))/(cos^2(2x))#

#y'=(2cos^2(2x)+2sin^2(2x))/(cos^2(2x))#

#y'=(2(cos^2(2x)+sin^2(2x)))/(cos^2(2x))#

To simplify this further we use the Pythagorean Identity:

#sin^2(x)+cos^2(x)=1#

So #2(cos^2(2x)+sin^2(2x))# becomes #2xx1=2#

Thus:

#y'=(2(cos^2(2x)+sin^2(2x)))/(cos^2(2x))#

becomes

#y'=(2)/(cos^2(2x))#

#=2sec^2(2x)#

There is a simple answer to this.

#d/(dx)[tanu] = sec^2u((du)/(dx))#

Thus, with #u = 2x# (thus, #u/x = 2#), #(du)/(dx) = 2#, and

#d/(dx)[tan(2x)] = sec^2(2x)*2 = color(blue)(2sec^2(2x))#

We have to find derivative of #color(blue)(y=tan(2x))#
Let us assume #u=2x#
Then
#color(blue)(y=tanu# & #color(blue)(u=2x#
By Chain Rule of Differenciation
#dy/dx=dy/(du)*(du)/(dx)#
#d/(dx)(tan2x)=(d/(du)(tanu))(d/dx(2x))#
#d/(dx)(tan2x)=##sec^2u*2 ->1#
Substitute u=2x in Eqn 1
Therefore
#d/dx(tan2x)=2sec^2 2x#