What is the derivative of #y=sin(1+tan(2x))#?

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Answer 1 · 2016-08-19T23:56:33

#2sec^2(2x)*cos(1+tan2x)#

Finding the derivative of this function by applying chain rule and using derivative of some trigonometric functions.

Let:
#u(x)=1+tan(2x)#
#f(x)=sinx#

Then #f(u(x))=sin(1+tan(2x))#

#f(u(x)))# is a composite function where its derivative can be found by applying chain rule that says:

#color(blue)(f(u(x))'=u'(x)*f'(u(x))#

Let's find #u'(x)# and # f'(x)#

#u'(x)=(2x)'*tan'(2x)#
#color(blue)(u'(x)=2sec^2(2x))#

#color(blue)(f'(x)=cosx)#

#(f(u(x)))'=u'(x)*f'(u(x))#

#(f(u(x)))'=2sec^2(2x)*cos(u(x))#

#color(red)((sin(1+tan2x))'=2sec^2(2x)*cos(1+tan2x))#