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derivative of 9[sin(x)]^2 is found by first letting u(x)=[sin(x)]^2. Note that sin2x = [sin(x)]^2, and the ^2 means raising the base to the exponent 2.

Find the d(9u(x))/dx using the chain rule.

d( 9u(x) )/dx = (d(9u)/du)(du/dx ) , by the chain rule.

So we need:

d(9u)/du = 9ulog(9)

du/dx = d( [sin(x)]^2 )/dx = 2sin(x) d( sin(x) )/dx = 2sin(x)cos(x)

Puttin this together gives:

d( 9u(x) )/dx = 9u(log(9)) 2sin(x)cos(x)

Now substitute in u(x) = [sin(x)]^2.

d( 9u(x) )/dx = 9[sin(x)]^2(log(9)) 2sin(x)cos(x)

= 2 log(9) 9[sin(x)]^2sin(x)cos(x) or

= log(9) 9[sin(x)]^2sin(2x)

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