How do you differentiate # y= sin2x-cos2x# using the chain rule?

1 Answer
Jan 31, 2016

Answer:

#dy/dx = 2cos2x+2sin2x#

Explanation:

The chain rule basically says that the derivative of a composite function (like #sin2x#) is the derivative of the main function times the derivative of the interior function. In the case of #sin2x#, we have the derivative of the main function (#sin2x# ) as #cos2x#, and the derivative of the interior function (#2x#) as #2#. So, by the chain rule, the derivative of #sin2x# is #2cos2x#.

The same logic applies to #cos2x#. The derivative of cosine is negative sine (#-sin#). so we begin with #-sin2x#. Then, because of the chain rule, we multiply this by the derivative of #2x#, which is, of course, #2#. Therefore, the derivative of #cos2x# is #-2sin2x#.

Finally we put it all together. We can express #dy/dx (sin2x-cos2x)# as #dy/dx (sin2x) - dy/dx (cos2x)#. Because we found the derivatives of these guys earlier, all we do is substitute:

#dy/dx (sin2x) = 2cos2x# and #dy/dx (cos2x) = -2sin2x#

#dy/dx = 2cos2x - (-2sin2x) = 2cos2x+2sin2x#