Find the first two derivatives of 2 sinx cosx.

We need to find the first derivative of 2sin(x)cos(x)


Let us assume y = 2sin(x)cos(x)

Use the product rule:

uv’ + vu’

where u is 2sin(x) and v is cos(x)

To find first derivative:

y’ = 2sin(x)(-sin(x)) + cos(x)2cos(x)

On simplifying we get

y’ = 2cos2(x)-2sin2(x)

y’ = 2(cos2(x)-sin2(x))

We know the trignometric identity

cos(2x) = cos2(x)-sin2(x)

Hence substituting in the above equation we get,

y’ = 2cos(2x)

First derivative of 2sin(x)cos(x) is 2cos(2x)

Now will find second derivative

Take second derivative using chain rule:

y” = 2(-sin(2x)cos(2x))


y” = -2sin(2x)(2)


y” = -4sin(2x)y” = -4sin(2x)


Hence the first two derivaties of 2sin(x)cos(x) are

y’ = 2cos(2x)

y” = -4sin(2x)

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