How do you differentiate #y=sin^(2)x - 2cosx#?

1 Answer
Kevin L.
Feb 26, 2017

#dy/dx = 2sin(x)cos(x) + 2sinx#

Explanation:

#sin^2x# can also be written as #(sinx)^2#. Bring down the two, and change the exponent to one. Also multiply it by the derivative of the inside. This gives you #2sin(x)cos(x)#.
The derivative of #2cos(x)# is #-2sin(x)#. Plug this into the equation and you get # + 2sin(x)#. When you add it altogether, you get #dy/dx = 2sin(x)cos(x) + 2sinx#

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