What is the derivative of #sin^2(x)#?

Answer

#2sin(x)cos(x)#

Explanation

You would use the chain rule to solve this. To do that, you'll have to determine what the "outer" function is and what the "inner" function composed in the outer function is.

In this case, #sin(x)# is the inner function that is composed as part of the #sin^2(x)#. To look at it another way, let's denote #u#=#sin(x)# so that #u^2#=#sin^2(x)#. Do you notice how the composite function works here? The outer function of #u^2# squares the inner function of #u=sin(x)#. Don't let the #u# confuse you, it's just to show you how one function is a composite of the other. Once you understand this, you can derive.

So, mathematically, the chain rule is:

The derivative of a composite function F(x) is:

F'(x)=f'(g(x))(g'(x))

Or, in words:

the derivative of the outer function (with the inside function left alone!) times the derivative of the inner function.

1) The derivative of the outer function(with the inside function left alone) is:

#d/dx u^2= 2u#
(I'm leaving the #u# in for now but you can sub in #u=sin(x)# if you want to while you're doing the steps. Remember that these are just steps, the actual derivative of the question is shown at the bottom)

2) The derivative of the inner function:

#d/dx sin (x) = cos (x)#

Combining the two steps through multiplication to get the derivative:

#d/dx sin^2(x)=2ucos (x)=2sin(x)cos(x)#