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How do you differentiate 9 to the power sin squared x sorry I wasn't allowed to use mathematical notation when asking?
Wiki User
∙ 12y ago
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)
Wiki User
∙ 12y ago
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